The Leading Maths - 10

Algebra 150 Allied The Leading Mathematics-10 Sequence and Series 151 6th mean (m6) = a + 6d = 20 – 6×4 = – 4; 7th mean (m7) = a + 7d = 20 – 7×4 = – 8 8th mean (m8) = a+8d = 20–8×4 = – 12; 9th mean (m9) = a + 9d = 20 – 9×4 = – 16 Example-6 There are n arithmetic means between 3 and 35. The ratio of the third mean and sixth mean is 5:9. (a) What is the common difference (d) in term of n ? (b) Find the values of n and d. (c) Find the fourth mean. Solution: (a) Here, the first term (a) = 3, the last term (b) = 35, the number of arithmetic means = n, The common difference (d) = b – a n + 1 = 35 – 3 n + 1 = 32 n + 1 . (b) By given, a + 3d a + 6d = 5 9 or, 3 + 3 × 32 n + 1 3 + 6 × 32 n + 1 = 5 9 or, 3n + 3 + 96 3n + 3 + 192 = 5 9 or, 27n + 27 + 864 = 15n + 15 + 960 or, 27n – 15n = 975 – 891 or, 12n = 84 => n = 7. ∴ d = 32 n + 1 = 32 7 + 1 = 32 8 = 4. (b) The required fourth mean (m4) = a + 4d = 3 + 4 × 4 = 19. Example-7 10 arithmetic means are inserted between a and b such that the third mean and eighth mean are 13 and 43 respectively. Find the values of a and b. Solution: Given, the first term = a, the last term = b, the number of arithmetic means (n) = 10, the common difference (d) = b – a n + 1 = b – a 10 + 1 = b – a 11 Now, the third mean (m3) = 13 or, a + 3d = 13 or, a + 3 × b – a 11 = 13 or, 10a + 3b – 3a 11 = 13 or, 8a + 3b = 143 ……….. (i) × 8 The eighth mean (m8) = 43 or, a + 8d = 43 or, a + 8 × b – a 11 = 43 or, 11a + 8b – 8a 11 = 43 or, 3a + 8b = 473 …..…… (ii) × 3 Multiplying eqn (i) by 8 and eqn (ii) by 3 and then subtracting them, we get

Algebra 152 Allied The Leading Mathematics-10 Sequence and Series 153 64a+24b = 1144 9a + 24b = 1419 55a + 0 = –275 or, a = – 5. Putting the value of a in eqn (i), we get 8 × (– 5) + 3b = 143 or, 3b = 143 + 40 or, 3b = 183 or, b = 61. ∴ a = – 5 and b = 61. PRACTICE 6.1 Read Think Understand Do Keeping Skill Sharp 1. (a) What is arithmetic mean ? (b) What is the arithmetic mean between p and q ? (c) What is the common difference of the sequence having the first term a and the last term b where there are n arithmetic means inserted between them ? (d) The arithmetic mean between two terms – 15 and 29 is ......... . (e) The 9th term of arithmetic sequence, whose 8th and 10th terms are 25 and 13 is ...... . 2. Circle ( ) the correct answer. (a) If b is the arithmetic mean between c and d of an AP, which is the correct relation between b, c and d ? (i) 2b = c + d (ii) 2c = b + d (iii) 2d = b + c (iv) None of them (b) Which is the arithmetic mean between 1 2 and 1 4 ? (i) 1 3 (ii) 3 4 (iii) 3 8 (iv) 2 3 (c) If there are 5 arithmetic means between 2 and 128, how many terms are there in it? (i) 16 (ii) 65 (iii) 5 (iv) 7 (d) If there are 3 arithmetic means between 5 and 25, which is the correct common difference ? (i) 4 (ii) 5 (iii) 15 (iv) 20 3 (e) If the common difference of an AP is 4 having the first term 7 and the last term 37, the 4th mean is .... . (i) 22 (ii) 23 (iii) 15 (iv) 11 – – –

Algebra 152 Allied The Leading Mathematics-10 Sequence and Series 153 Check Your Performance 3. (i) The two numbers 7 5 and 5 2 are the first and the last terms of an arithmetic sequence (AS). (a) What is an arithmetic mean (AM) between two terms ? Define it. (b) Find the AM of the given two numbers. (c) What is the 10th term of the AS ? Find it. (ii) The 15th term and the 17th term of an AP of n terms are 100 and 10 respectively. (a) Find the 16th term of the sequence. (b) Find the common difference of the sequence. (c) What is its last term ? Find it. 4. (i) 10 is the AM between 4x + 1 and 7x – 3. (a) Find the value of x. (b) Arrange these three terms in the arithmetic sequence. (c) How much percent is increase in each term in the AS ? (ii) 2p2 – 1, 2p + 1 and 3p are in an AP. (a) Find the three terms. (b) Write the next three terms containing in the same arithmetic sequence. (c) Compute the general term of the sequence. (iii) Two numbers are in the ratio 3:7. If their AM is 50, find the two numbers. (iv) The arithmetic mean between two terms is 13 5 . If the first term is 7 6 , find the next term. 5. (i) If x, 4, y, 32 are in AP, write the values of x and y. (a) What is y in the AP ? (b) Find the value of x. (c) Write the relation between x and y. (ii) Find the values of p and q when 2 1 2 , p, 221 2, q are in an arithmetic sequence. (a) What is p in the AP ? (b) Find the value of q. (c) What is the common difference of the AP? 6. (ii) There are n-arithmetic means between two terms p and q of an AP. (a) Write the total number of terms in the AP. (b) What is the common difference of the AP? Find it. (c) If the values of p and q are – 2 and 16 respectively, insert five arithmetic means between them. (ii) Find the values of a, b and c if 19, a, b, 40, c are in AP. 7. (i) There are n-arithmetic means between – 25 and 10. If its third mean is – 10, (a) find the value of n.

Algebra 154 Allied The Leading Mathematics-10 Sequence and Series 155 (b) How many total terms are there in the AP ? (c) Find the fifth mean. (ii) There are k-AM’s between 1 and 4. If the third mean is 14 5 . (a) Find the value of k. (b) How many total terms are there in the AP ? (c) Find the other remaining means. (iii) There are five arithmetic means between 5 and b. The last mean is 25. (a) Find the value of b. (b) Find the other remaining means. 8. (i) There are k-arithmetic means between 25 and 49. The ratio of the third mean and fifth mean is 17:20. (a) What is the common difference (d) in term of n ? (b) Find the values of n and d. (c) Find the sixth mean. (ii) There are m-arithmetic means between 175 and 125 such that m1:mn=17:13. (a) What is the common difference (d) in term of n ? (b) Find the values of n and d. (c) Find the sixth mean. 9. (i) 6 arithmetic means are inserted between two terms a and b such that the third mean and the sixth mean are 142 and 154 respectively. (a) Find the values of a and b. (b) Find the sequence. (ii) 10 arithmetic means are inserted between any two numbers such that the fourth and ninth means are 1 and – 29 respectively. Find the two numbers. 3. (i) (b) 1 19 20 (c) 6 7 20 (ii) (a) 55 (b) – 45 (c) 775 – 45n 4. (i) (a) 2 (b) 9, 10, 11 (c) 22 2 9 % (ii) (a) 1, – 1, – 3 or 7 2 , 4 , 2 9 % (b) – 5, – 7, – 9 or 5, 11 2 , 6 (c) 3 – 2n or 4 + 3n 2 (iii) 30, 70 (iv) 121 30 5. (i) (a) 18 (b) – 10 (c) x + 28 = y (ii) (a) 25 2 (b) 65 2 (c) 10 6. (i) (a) n + 2 (b) q – p n + 1 (c) 1, 4, 7, 10, 13 (ii) 26, 33, 47 7. (i) (a) 6 (b) 8 (c) 0 (ii) (a) 4 (b) 6 (c) 8 5 , 11 5 , 17 5 , 4, 23 5 (iii) (a) 45 (b) 9, 13, 17, 21 8. (i) (a) 24 n + 1 (b) 7, 3 (c) 43 (ii) (a) 700 13 – n (b) 9, – 5 (c) 145 9. (i) (a) 130, 158 (b) 130, 134, 138, 142, ... (ii) 25, – 41 Answers

Algebra 154 Allied The Leading Mathematics-10 Sequence and Series 155 6.2 Sum of Arithmetic Series At the end of this topic, the students will be able to: ¾ solve the verbal problems related to arithmetic series. Learning Objectives I Introduction Activity 1 Nerisha puts Rs. 10 in the money saving pot on the first day, Rs. 15 on the second day, Rs. 20 on the third day, Rs. 25 on the fourth day and Rs. 30 on the fifth day. How much money does she save in the pot ? By adding each amount, we obtain the sum 100. i.e., 10 + 15 + 20 + 25 + 30 = 100. This is the sum of the first five terms of the arithmetic series. The numerical sum of every term of an arithmetic series is called the sum of the AP. The sum of the first n terms of the arithmetic series is denoted by Sn. How can we take the sum easily by using formula? For this the sum of the first five terms of the given AP, S5 = 10 + 15 + 20 + 25 + 30 ……….… (i) Writing this series in the reverse order, we get S5 = 30 + 25 + 20 + 15 + 10 ………….. (ii) Adding the relations (i) and (ii), we get 2S5 = 40 + 40 + 40 + 40 + 40 or, 2S5 = 5 × 40 or, 2S5 = 200 ∴ S5 = 100. Similarly, let a be the first term, d the common difference, n the number of terms and l the last term of a series in AP. Suppose the sum of the first n terms is denoted by Sn. ∴ Sn = a + (a + d) + (a + 2d) + …….. + (l – 2d) + (l – d) + l ………… (i) Writing the series in the reverse order, we get Sn = l + (l – d) + (l – 2d) + …...... + (a + 2d) + (a + d) + a …….....…. (ii) Adding the corresponding terms of eqn (i) and eqn (ii), we obtain 2Sn = (a + l) + (a + l) + (a + l) +…… + (a + l) + (a + l) + (a + l) or, 2Sn = n(a + l) ∴ Sn = n(a + l) 2 But, we have; l = a + (n – 1)d ∴ Sn = n 2 [2a + (n – 1)d]

Algebra 156 Allied The Leading Mathematics-10 Sequence and Series 157 Let us derive the formula of the sum of the arithmetic series as method given below; This formula was invented by a German Mathematician Johann Carl Friedrich Gauss (1777–1855) during his school time by computing sum of the integers from 1 to 100 as shown in the above right chart. II Sum of Natural Numbers (i) Sum of the First n-Natural Numbers What is the sum of angles in the digits 1 to 9 ? They form an arithmetic series as 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9. We can easily calculate, which is 45. How? By which method do we use for the sum of natural numbers that are starting from 1 ? Suppose, S9 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 and S9 = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 By adding, 2S9 = 10+10+10+10+10+10+10+10+10 or, S9 = 90 2 = 9 × 10 2 = 9(9 + 1) 2 The series of the first n-natural numbers is given. Its sum will be n(n + 1) 2 . Now, take a series of the natural numbers, 1 + 2 + 3 + 4 + 5 + ………. to n terms. The first term (a) = 1, The common difference (d) = 1, The number of terms (n) = n Now, we have The sum of the first n terms (SN) = n 2 [2 × 1 + (n – 1) × 1] = n 2 [2 + n – 1] = n(n + 1) 2 Hence, the sum of the series of the first n natural numbers is SN = n(n + 1) 2 . (ii) Sum of the First n-natural Even Numbers How many total legs of the adjoining peasant, sheep and goats that are arranged in arithmetic sequence as 2, 4, 6, 8, 10? There are 30 legs. How do we derive the rule for the series of even numbers? The first n-even numbers are 2, 4, 6, 8, 10,……….., 2n. Addition of 101 up to 50 times 50 101 = 5050 101 101 101 101 Trick ! Just for It. 1) 1 + 2 + 3 + ... + 20 = (1 + 20) × 10 = 210 , , , ,

Algebra 156 Allied The Leading Mathematics-10 Sequence and Series 157 Then its sum is (SE) = 2 + 4 + 6 + 8 + 10 + …………. + 2n = 2(1 + 2 + 3 + 4 + 5 + ……. + n) = 2 n(n + 1) 2 = n(n + 1). Hence, the sum of the series of the first n-even number is SE = n (n + 1). (iii) Sum of the First n-natural Odd Numbers The first n-odd natural numbers are 1, 3, 5, 7, 9,…………, 2n – 1. Then, its sum, Sn= 1 + 3 + 5 + 7 + 9 + ……… + (2n – 1) Adding 1 + 1 + 1 + ………. + 1 (n terms) on both sides, we get Sn+ (1 + 1 + 1 + .......... + 1) = 1 + 3 + 5 + 7 + 9+ .......... + (2n – 1) + (1 + 1 + 1 + 1 + 1 + ..........+ 1) or, Sn+ n = (1 + 1) + (3 + 1) + (5 + 1) + (7 + 1) + (9 + 1) +……….+ (2n – 1 + 1) or, Sn+ n = 2 + 4 + 6 + 8 + 10 +………+ 2n or, Sn+ n = n(n + 1) or, Sn+ n = n2 + n ∴ Sn = n2 . Hence, the sum of the series of the first n-natural odd numbers is SO = n2 . Alert to Arithmetic Sum (i) Arithmetic series is not only addition form, it would be in subtraction form. (ii) The sum of natural and whole numbers are not different. Points to be Remembered 1. The sum or additional form of an arithmetic progression is called an arithmetic series. 2. The sum of the 1st n-terms of an arithmetic series is calculated by the formula Sn = n(a + l) 2 or Sn = n 2 [2a + (n – 1)d]. 3. The sum of the first n-natural numbers is calculated by the formula SN = n(n + 1) 2 . 4. The sum of the first n-even numbers SE = n(n + 1) 2 . 5. The sum of the first n-odd numbers SO = n2 . Example-1 The arithmetic series is 1 + 4 + 7 + ..... + 31. (a) Which formula is used to find the sum of an arithmetic series ? (i) n 2(a + l) (ii) n 2[2a + (n + 1)d] (iii) n 2[2a + (n – 1)d] (iv) Both (i) or (iii) (b) Find the common difference of the given series. (c) Find the sum of the given series. (d) If 6 next terms are added in the given series, what will be its sum ? Find it. Solution: (a) Here, (i) n 2(a + l) or (iii) n 2[2a + (n – 1)d] is used to find the sum of an arithmetic series. Oh! We can compute these relations by using sum formula: Sn = n 2 [2a + (n – 1)d]

Algebra 158 Allied The Leading Mathematics-10 Sequence and Series 159 (b) Common difference of the given series (d) = 4 – 1 = 3. (c) In the given series, the first term (a) = 1, the last term (l) = 31, No. f terms (n) = 10 Now, we have Sum of the first 10 terms of the arithmetic series (S10) = n 2(a + l) = 10 2 (1 + 31) = 10 2 × 32 = 160. (d) If 6 next terms are added in the given series, the total number of terms (N) = 10 + 6 = 16 ∴ Sum of the first 16 terms of the arithmetic series (S10) = N 2[2a + (N – 1)d] = 10 2 [2 × 1 + (16 – 1)3] = 5[2 + 45] = 235. Example-2 (a) What is the sum of odd numbers from 1 to 50 ? (b) What is the sum of even numbers 2 + 4 + 6 + ........ up to 50 ? (c) Which sum is more among the sum of odd and even numbers from 1 to 50 and by how much? Why ? Write their relation. Solution: (a) Number of odd numbers from 1 to 50 (n) = 50 2 = 25. ∴ Sum of the first 25 odd numbers from 1 to 50 (SO) = n2 = 252 = 625. (b) Number of even numbers 1 + 3 + 5 + ........ up to 50 (n) = 50 2 = 25. ∴ Sum of the first 25 even numbers from 1 to 50, SE = n(n + 1) = 25(25 + 1) = 650 (c) SE is more than SO by 650 – 625 = 25 because each number of the even series is 1 more than each term of odd series for all 25 terms in each series. The relation between SE and SO is SE = SO + 25. Example-3 nth term of an AP is 2n + 5. (a) Find the arithmetic series having nth term mentioned above. (b) Find its common difference. (c) Find the sum of the first 30 terms of an AP using common difference. (d) Find the sum of the first 30 terms of an AP using last term. Solution: (a) Given, the nth term of an AP (tn) = 2n + 5 Now, the first term (t1) = 2 × 1 + 5 = 7, the second term (t2) = 2 × 2 + 5 = 9, the third term (t3) = 2 × 3 + 5 =11 and so on ∴ The required arithmetic series is 7 + 9 + 11 + ....... (b) Common difference (d) = 7 – 5 = 2,

Algebra 158 Allied The Leading Mathematics-10 Sequence and Series 159 (c) Number of terms (n) = 30, Sum of the first 30 terms (S30) = ? Now, Sum of the first 30 terms, S30 = n 2[2a + (n – 1)d] = 30 2 [2 × 7 + (30 – 1)2] = 15(14 + 58) = 15 × 72 = 1080. (d) The 30th term or last term (l) = 2 × 30 + 5 = 65, Sum of the first 30 terms (S30) = ? Now, Sum of the first 30 terms, S30 = n 2(a + l) = 30 2 (7 + 65) = 15 × 72 = 1080. Example-4 The sum of the first nine terms in an AS with the first term 20 is 0. (a) Find the common difference. (b) Find the sum of its first 20 terms. Solution: (a) Given, in an AP, the first term (a) = 20, the number of terms (n) = 9, the sum of the first nine terms (S9) = 0, the common difference (d) = ? Now, we have Sn = n 2[2a + (n – 1)d] or, 0 = 9 2[2 × 20 + (9 – 1)d] or, 0 = 40 + 8d or, 8d = – 40 ∴ d = – 5 Hence, the required common difference of the AP is – 5. (b) Sum of the first 20 terms (S20) = 20 2 [2 × 20 + (20 – 1)(– 5)] = 10[40 – 95] = 10(– 55) = – 550. Example-5 The 10th term of an AS is 23. (a) Find the sum of the first 19 terms. (b) If you suppose any number for the first term a, prove that the sum of its 19 terms is the same in the no. (a). (c) If you suppose any number for the common difference d, prove that the sum of its 19 terms is the same in the no. (a) or (b). Solution: (a) Here, in an AP, the 10th term (t10) = 23 or, a + 9d = 23 ............ (i) Now, Sum of the first 19 terms (S19) = 19 2 [2a + (19 – 1)d] = 19 2 × 2[a + 9d] = 19 × 23 [∴ From relation (i)] = 437.

Algebra 160 Allied The Leading Mathematics-10 Sequence and Series 161 (b) Suppose, the first term (a) = 50 From the relation (i), 50 + 9d = 23 => 9d = 23 – 50 ∴ d = – 27/9 = – 3 ∴ Sum of the first 19 terms (S30) = 19 2 [2 × 50 + (19 – 1) (– 3)] = 19 2 [100 – 54] = 19 2 × 46 = 437. Proved. (c) Suppose, common difference (d) = 8 From the relation (i), a + 9 × 8 = 23 => a = 23 – 72 ∴ a = – 49 ∴ Sum of the first 19 terms (S30) = 19 2 [2 × (– 49) + (19 – 1) (8)] = 19 2 [– 98 + 144] = 19 2 × 46 = 437. Proved. Example-6 The sum of a series in AP with the first term 3 is 1395 and its last term, 90. (a) Find the number of terms. (b) Find its common difference. (c) Find the sum of its 20 terms. Solution: (a) Given, the first term (a) = 3, sum of AP (Sn) = 1395, the last term (l) = 90 Number of term (n) = ?, Common difference (d) = ? Now, Sn = n 2 (a + l) or, 1395 = n 2 (3 + 90) or, 2790 = 93n or, n = 30 (b) l = a + (n – 1)d or, 90 = 3 + (30 – 1)d or, 90 – 3 = 29d or, 29d = 87 or, d = 3 (c) Again, the first term (a) = 3, common difference (d) = 3, number of term (n) = 20 ∴ Sum of the first 20 terms (S30) = 20 2 [2 × 3 + (20 – 1) (3)] = 10[6 + 57] = 630. Example-7 The sum of the first twelve terms of an AP is 294 and its last term exceeds the double of its fifth term by 1. (a) Find its first term and common difference. (b) Find the sum of its first 20 terms. Solution: (a) Given, in AP, let a and d be the first term and common difference. Sum of the first twelve terms (S12) = 294 or, 12 2 [2a + (12 – 1) d] = 294 or, 2a + 11d = 49 ................. (i) Again, the last term (l) = 2 × the fifth term (t5) + 1 or, a + (12 – 1)d = 2 × (a + 4d) + 1 or, a + 11d = 2a + 8d + 1 or, a – 3d = – 1 .………….. (ii)

Algebra 160 Allied The Leading Mathematics-10 Sequence and Series 161 Multiplying by 2 in eqn (ii) and subtracting with the eqn (i), we get 2a + 11d = 49 –2a 6d = 2 17d = 51 ∴ d = 3. (b) Sum of its first 20 terms (S20) = 20 2 [2 × 8 + (20 – 1)3] = 10(16 + 19 × 3) = 730. Example-8 If the sum of the first 4 terms of an arithmetic series (AS) is 158 and that of the first 10 terms is 485. (a) Find its first term and common difference. (b) Find the 20th term of the AS. Solution: (a) Given, in an AP, Sum of the first 4 terms (S4) = 158 or, 4 2 [2a + (4 – 1)d] = 158 or, 2(2a + 3d) = 158 or, 2a + 3d = 79 ....... (i) Again, Sum of the first 10 terms (S10) = 485 or, 10 2 [2a + (10 – 1)d] = 485 or, 5(2a + 9d) = 485 or, 2a + 9d = 97 ….. (ii) Subtracting the eqn (i) from the eqn (ii), then 6d = 18 => d = 3. Putting d = 3 in eqn (i), we obtain, 2a + 3 × 3 = 79 or, 2a = 79 – 9 or, 2a = 70 => a = 35. Example-9 The sum of three consecutive numbers in an AP is 60 and the sum of their square is 1400. (a) What is the middle number ? (b) Find the numbers in sequential form. Solution: (a) Suppose, a – d, a and a + d are in AP, then a – d + a + a + d = 60 or, 3a = 60 ∴ a = 20. (b) Again, (a – d)2 + a2 + (a + d)2 = 1400 or, a2 – 2ad + d2 + a2 + a2 + 2ad + d2 = 1400 or, 3a2 + 2d2 = 1400 or, 3 × (20)2 + 2d2 = 1400 or, 1200 + 2d2 = 1400 Putting the value of d in eqn (ii), we get, a – 3 × 3 = – 1 or, a = 9 – 1 = 8. (b) Now, the required 20th term (t20) = a + (n – 1)d = 35 + (20 – 1)3 = 35 + 57 = 92

Algebra 162 Allied The Leading Mathematics-10 Sequence and Series 163 or, d2 = 1400 – 1200 2 = 100 ∴ d = ± 10. If a = 20 and d = 10, then the required numbers are (20 – 10), 20, (20 + 10) = 10, 20, 30. If a = 20 and d = –10, then the required numbers are (20 + 10), 20, (20 – 10) = 30, 20, 10. Example-10 A 55 year old man has 5 sons born at equal intervals. The sum of the ages of the father and his sons is 130 years. (a) Find the ages of all the youngest sons while the eldest son is 25 years. (b) What percent of age is more of the elder son than the youngest son ? Calculate. Solution: (a) Here, Age of the man (father) = 55 years, Sum of ages of the father and his 5 sons = 130 yrs, Age of 5 sons (Sn) = 130 – 55 = 75 years, Age of the eldest son (l) = 25 years, Age of the youngest son (a) = ? Now, we have Sn = n 2 (a + l) or, 75 = 5 2 (a + 25) or, 75 × 2 5 = a + 25 or, 30 – 25 = a or, a = 5. Again, Common difference of age (d) = l – a n – 1 = 25 – 5 5 – 1 = 20 4 = 5 ∴ The ages of all the youngest sons are 5 yrs, 5 + 5 = 10 yrs, 10 + 5 = 15 yrs; and 15 + 5 = 20 yrs. (b) The required more percent of the age of elder son than the age of youngest son = 25 – 5 5 × 100% = 20 5 × 100% = 4 × 100% = 400%.

Algebra 162 Allied The Leading Mathematics-10 Sequence and Series 163 PRACTICE 6.2 Read Think Understand Do Keeping Skill Sharp 1. (a) Write the sum of the arithmetic series a + (a + d) + (a + 2d) + ...... up to n terms. (b) What is the sum of the arithmetic series of n terms having the first term a, the last term l? (c) Write the sum of the first 10 terms of an AP having the first term 5 and the last term 50. 2. (a) The formula to find the sum of the first n-even numbers is ........... . (b) The sum of natural numbers up to 20 is ........... . (c) The sum of the first 10 odd numbers is ........... . 3. Circle ( ) the correct answer. (a) What is the sum of the first n-terms of an arithmetic series with the first term a and common difference d? (i) Sn = n 2 [2a + (n + 1)d] (ii) Sn = n 2 [2a + (n – 1)d] (iii) Sn = a 2 [2a + (n – 1)d] (iv) Sn = n 2 [2a + (d – 1)n] (b) What is the sum of the first n-even numbers ? (i) 2n (ii) n2 (iii) n(n + 1) (iv) n(n + 1) 2 (c) What is the sum of the series 1 + 3 + 5 + 7 + ........ up to 20 terms ? (i) 400 (ii) 200 (iii) 210 (iv) 420 (d) The sum of the first 200 natural numbers is ........ . (i) 40000 (ii) 40200 (iii) 400 (iv) 20100 (e) Which is the correct sum of the arithmetic series 4 + 10 + 16 + 22 + .... + 118 having 20 terms ? (i) 1220 (ii) 2440 (iii) 7021 (iv) 14042

Algebra 164 Allied The Leading Mathematics-10 Sequence and Series 165 Check Your Performance 4. Answer the following questions for the given series. (i) – 20 – 13 – 6 + …………. to 15 terms. (ii) 9 + 33 4 + 15 2 + 27 4 + ……. to 25 terms. (iii) 4 + 9 + 14 + 19 + ………. + 54 (iv) 0.33 + 0.36 + 0.39 + 0.42 + …. + 0.99 (a) Find the common difference of the given series. (b) Find the sum of the given series. (c) If 10 next terms are added in the given series, what will be its sum ? Find it. 5. (a) What is the sum of odd numbers from 1 to 100 ? (b) What is the sum of even numbers 2 + 4 + 6 + ........ to 100 terms ? (c) Which sum is more the sum of odd numbers or even numbers from 1 to 100 and by how many? Why? Write their relation. 6. (a) Why is the formula of the sum of natural number different from that of even numbers ? (b) Write the sum of the following natural numbers. (i) 1 + 2 + 3 + 4 + ………… to 30 terms. (ii) 2 + 4 + 6 + 8 + ……...…. to 50 terms. (c) Find the sum of the first nth terms of the following series in AP: (i) 6 + 11 + 16 + …………… (ii) 3 2 3 + 41 3 + 5 + ………… 7. nth term of an APs are given below. (i) tn = 2n + 1 (ii) an = 20 – 3n (a) Find the arithmetic series having the nth term mentioned above. (b) Find its common difference. (c) Find the sum of the first 25 terms of an AP using common difference. (d) Find the sum of the first 25 terms of an AP using the last term. 8. (i) The sum of the first 16 terms of an arithmetic series with the first term 10, is 520. (a) Find the common difference. (b) Find the sum of its first 25 terms. (ii) The sum of the first 6 terms of AP is 2166 and its common difference is 104. (a) Find the first term of the AP. (b) Find the sum of its first 20 terms.

Algebra 164 Allied The Leading Mathematics-10 Sequence and Series 165 (iii) The sum of a series in AP is –13 whose first term and common difference are 35 and – 6 respectively. (a) Find the number of terms in the series. (b) Find the sum of its first 30 terms. 9. The 11th term of an AS is 23. (a) Find the sum of the first 21 terms. (b) If you suppose any number for the first term a, prove that the sum of its 21 terms is the same in the no. (a). (c) If you suppose any number for the common difference d, prove that the sum of its 21 terms is the same in the no. (a) or (b). 10. (i) The first and the last terms of the arithmetic series are 53 and – 31 respectively. The sum of the series is 165. (a) Find the number of terms. (b) Find their common difference. (c) Find the sum of its 25 terms. (ii) The first term and common difference in the AP are 3 and 4 respectively. (a) How many terms must be taken to make its sum 210 ? (b) Find its last term. 11. (i) The sum of the first 10 terms of an AP is 185 and its last term is 4 times of the second term. (a) Find its first term and common difference. (b) Find the sum of its first 16 terms. (ii) The sum of the first 10 terms of an arithmetic series is 145 and the sum of the fourth term and the tenth term is 20. (a) Find its first term and common difference. (b) Find the sum of its first 15 terms. (iii) The sum of the first 10 terms of a series in an AP is 365 and its eighth term exceeds the double of the fourth term by 2. (a) Find its first term and common difference. (b) Find the sum of its first 15 terms. 12. (i) A woman of 45 years old has 8 children born at equal intervals of year. The sum of the ages of the mother and her children is 145 years. (a) What is the age of the eldest child, if the youngest child is 2 years old?

Algebra 166 Allied The Leading Mathematics-10 Sequence and Series 167 (b) How many times the age is more of the elder child than the youngest child ? Calculate. (ii) 100 stones are placed in a straight line, exactly a meter apart, the first is one meter from the basket. What distance will a person walk, who gathers single stone to put in the basket and return to collects another and continue in this way up to 100 times? (iii) Shova saves Rs. 20 during the first month, Rs. 30 in the next month and Rs. 40 in the third month. (a) If she continues to save money in this sequence, find the total money at the end of 2 years without interest. (b) In how many years will she save Rs. 18900 ? Find it. (iv) A man undertakes to pay off a debt of Rs. 622500 by monthly installments; he pays Rs. 10000 in the first month and continually increases the installments in every subsequent month by Rs. 100. In what time will the debt be cleared up? (v) The sum of Rs. 1000 is deposited every year at 8% simple interest. Find the interest at the end of each year. Do these interest amounts form an AP? If so, find the total interest at the end of 30 years. (vi) A gardener plans to construct a trapezoidal shaped structure in his garden. The longer side of trapezoid needs to start with a row of 97 bricks. Each row must be decreased by 2 bricks on each end and the construction should stop at 25th row. How many bricks does he need to buy ? 4. (i) (a) 7 (b) 58 (c) 128 (ii) (a) – 3 4 (b) 0 (c) – 15 2 (iii) (a) 5 (b) 319 (c) 609 (iv) (a) 0.03 (b) 15.18 (c) 21.78 5. (a) 250 (b) 10100 6. (b) (i) 465 (ii) 2550 (c) (i) 5n2 + 7n 2 (ii) n2 + 10n 3 7. (i) (a) 3 + 5 + 7 + 9 + ... (b) 2 (c) 675 (d) 675 8. (i) (a) 3 (b) 1150 (ii) (a) 10 (b) 21780 (iii) (a) 13 (b) – 1560 9. (a) 483 10. (i) (a) 15 (b) – 6 (c) 275 (ii) (a) 10 (b) 39 11. (i) (a) 5, 3 (b) 440 (ii) (a) 28, – 3 (b) 105 (iii) (a) 5, 7 (b) 1430 12. (i) (a) 23 years (b) 10.5 times (ii) 10100 meters (iii) Rs. 3240, 5 years (iv) 50 months (v) Rs. 37200 (vi) 1225 Bricks Answers

Algebra 166 Allied The Leading Mathematics-10 Sequence and Series 167 6.3 Mean/s of Geometric Sequence At the end of this topic, the students will be able to: ¾ solve the problems related to the means of geometric sequence. Learning Objectives I Introduction Activity 1 In six months, Kirana Stores A and B sell the following amount of goods. Name\Month January February March April May June A Rs. 4000 Rs. 8000 Rs. 16000 Rs. 32000 Rs. 64000 Rs. 128000 B Rs. 5000 Rs. 10000 Rs. 20000 Rs. 40000 Rs. 80000 Rs. 160000 Observe the above sequences of amounts and discuss on the following questions. ► What is the ratio of sales amount in consecutive two months of the Kirana Store A ? ► What is the ratio of sales amount in consecutive two months of the Kirana Store B ? ► Are the ratios of the sales amounts in consecutive two months of the Kirana Stores equal? ► What is the relation between the sales amount in consecutive three months for both stores ? From the above sales amount of the Kirana Store A, (i) 4000, 8000, 16000 (ii) 4000, 8000, 16000, 32000 (iii) 4000, 8000, 16000, 32000, 64000 Mean Means Means The term/s between the given two terms in a GP is /are called geometric mean/s (AM/s). The geometric mean was first invented by ancient Greek philosopher and mathematician Pythagoras of Samos (570 - 495 BC) and his students at the Pythagorean School of Mathematics in Cortona, a coastal city in ancient Greece on the work of Pythagoras' theorem. So, the geometric mean also is called Pythagorean mean. The geometric mean is used in interest rate, population growth and value depreciation, portfolio returns, stock indexes, etc. II Geometric Mean (GM) What is a number between two numbers 2 and 8 such that these numbers form a geometric sequence? A number between 2 and 8 is 4. But, how would it be? The number in betweenness number is 2 × 8 = 16 = 4. It is the square root of the product of the given two terms. Similarly, what are only one exactly middle numbers between the following set of two numbers such that the ratios of their successive terms are constants: 1, ......, 4; 4, ......, 9; 125, ......, 5; 8, ......, 128. Pythagorean School

Algebra 168 Allied The Leading Mathematics-10 Sequence and Series 169 Now, consider three terms a, g, b of GP. Their common ratios of successive two terms are equal. i.e., g a = b g or, g2 = ab ∴ g = ab. i.e. GM = ab . Hence, the GM between a and b is ab. Note : If tn-2, tn-1, tn are in GP, then tn -1 = tn – 2 × tn . III n-Geometric Means (GMs) Consider the following two numbers as the first and the last terms in a GP; 2, ___ , ___ , ___ , ___ , 64 and 1, ___ , ___ , ___ , ___ , ___ , ___ , ___ , 6561. How many geometric means are there between the given two numbers? What are they? The first is easy to find. But in the second, it is a little difficult. How can we easily find them? For removing these types of difficulties, we learn the rule that help to find the geometric means between any two numbers as follows. Suppose the GP has n-geometric means as g1, g2, g3, g4, ..., gn between two numberer a and b. Then the GP a, g1, g2, g3, g4, ....., gn, b has the total number of terms, N = n + 2. Now, we have the general terms of GP, (tn) = arN-1 or, b a = r N – 1 ∴ r = b a 1 N – 1 , where N = total no. of terms. i.e., r = b a 1 n + 2 – 1 or, r = b a 1 n + 1, where n = total no. of means. The n-geometric means are in the table below: Terms t1 t2 t3 t4 ……. t n-3 t n-2 tn-1 tn Means -- g1 g2 g3 ……. g n-2 g n-1 gn -- Formulae a ar = a b a 1 n + 1 ar2 = a b a 2 n + 1 ar3 = a b a 3 n + 1 ……. arn-2 = a b a n – 2 n + 1 arn-1 = a b a n – 1 n + 1 arn = a b a n n + 1 b Alert to Geometric Mean/s (i) GM has no different ratio from the first and the last terms. (ii) In GMs between any two numbers, all the ratios of any two consecutive terms are not different. IV Relation Between Arithmetic Mean and Geometric Mean The arithmetic progression has constant difference between any two consecutive terms (non-negative real numbers) whereas the geometric progression has constant ratio between any two successive terms. The AP and GP are two different progressions, but they are mathematically related by their means. Relation Between AM and GM of Two Numbers Let ‘a’ and ‘b’ be two positive or negative numbers then the arithmetic mean (AM) = a + b 2 and the geometric mean (GM) = ab.

Algebra 168 Allied The Leading Mathematics-10 Sequence and Series 169 Now, we take AM – GM = a + b 2 – ab = a + b – 2 ab 2 = ( a) 2 – 2 a . b + ( b) 2 2 = ( a – b) 2 2 Since ( a – b) 2 is non-negative, so AM – GM > 0. ∴ AM ≥ GM Note : (i) If a ≠ b, then AM > GM. (ii) If a = b, then AM = GM. Geometric Interpretation of AM-GM Inequality Draw a square ABCD and take the points E, F, G and H such that AE = BF = CG = DH = a and EB = FC = GD = HA = b in which a > b. Also, draw the rectangles based on a and b named as AEQH, BERF, CFSG and DGPH as shown in the adjoining diagram and the remaining inner part is a square PQRS, whose area is not zero. If a = b, the square PQRS will be zero. Then, AB = BC = CD = DA = a + b. So, Area of Sq. ABCD (A1) = (a + b) 2 Area of four rectangles (A2) = 4ab; Now, Area of Sq. ABCD (A1) > Area of four rectangles (A2) or, (a + b) 2 > 4ab or, ( a + b 2 )2 > ab ∴ a + b 2 > ab But, we know that AM of a and b (AM) = a + b 2 and GM between a and b (GM) = ab . Hence, AM > GM. Alert to Mixed Progression GM is never greater than AM or AM is never smaller than GM. Points to be Remembered 1. The term/s between the given two terms in an GP is/are called a geometric mean/s (AM/s). 2. GM between two numbers a and b is ab. i.e., GM = ab. 3. The common ratio of any two terms when inserting n-geometric means between a and b, is, r = b a 1 N – 1 ,where N = total no. of terms OR r = b a 1 n + 1, where n = total no. of geometric means. 4. The value of n-geometric means between a and b is arn , n = Numbers of means and n = 1, 2, 3, ... 5. AM is always greater or equal to GM for any two numbers. i.e., AM > GM or GM < AM. a a a a b b A b b B D C R Q P S R Q P S E F G H AM > GM can be shown by using semi-circle. A G P R ≥ a O Q b ab a + b 2

Algebra 170 Allied The Leading Mathematics-10 Sequence and Series 171 Example-1 (a) Define geometric mean (GM) between two numbers. (b) Find the GM between 5 and 245. (c) What is its 9th term ? Find it. Solution: (a) A middle term that has constant ratio with previous term and from forward term, is called geometric mean (GM) between two numbers. If a and b are two terms, the GM between them is ab. (b) The geometric mean between 5 and 245 is ab = 5 × 245 = 1225 = 35. (c) Now, in the given GS, first term (a) = 5, common ratio (r) = 35 5 = 7. No. of terms up to 9 terms (n) = 9, 9th term (t9) = ? We know that General term (tn) = arn – 1 or, t12 = 5 × 79 – 1 = 5 × 78 = 28824005. Example-2 18 is the geometric mean between two numbers 3x and 4x. (a) Find the value of x. (b) Add next two terms in the AS. (c) How much percent is the increased in each term in the AS ? Solution: (a) Given, since 18 is the GM between 3x and 4x so, 18 = 3x × 4x. By squaring on both sides, we get 324 = 12x2 or, 324 12 = x2 or, 27 = x2 or, x = 27 or, x = 9 × 3 or, x = 3 3. Thus, the value of x is 3 3. (c) The increased percent in each term in the GS = r a×100% = 2 9 3 × 3×100% = 7.41%. Example-3 Find the values of the x and y, when 1 8 , x, 8, y are in GP. (a) What is x in the GP ? (b) Find the values of x and y. (c) Write the relation between x and y. Solution: (a) x is a geometric mean between 1 8 and 8 of the given GP. (b) Given, 1 8 , x, 8, y are in GP. So, x = 1 8 × 8 = 1 = 1 [ GM = ab] ∴ x = 1 Again, 8 is the GM between x and y. So, 8 = xy or, 8 = 1 × y ∴ 64 = 1 × y or, y = 64 (b) Now, 1st term = 3x = 3 × 3 3 = 9 3, 3rd term = 4x = 4 × 3 3 = 12 3 ∴ Common ratio (r) = t2 ÷ t1 = 18 ÷ 9 3 = 2/ 3 ∴ The next two terms are 4th term = t3 × r = 12 3 × 2/ 3 = 24, 5th term = t4 × r = 24 × 2/ 3 = 16 3.

Algebra 170 Allied The Leading Mathematics-10 Sequence and Series 171 “Alternatively for p and q” Since 1 8, x, 8, y are in AP. So, x 1/8 = 8 x = y 8 Taking the first and the second parts, we get x 1/8 = 8 x or, 8x2 = 8 or, x2 = 1 ∴ x = 1. (c) The relation between x and y is y = xr2 , where r = 8. Example-4 There are n-geometric means between two terms a and b of an GP. (a) How many terms are there in the GP ? (b) What is the common ratio of the GP in term of n? Find it. (c) The values of a and b are 1 2 and 16 respectively. If the third mean is 4, find the number of means between these two numbers (d) Find the last mean. Solution: (a) There are n + 2 terms in the GP. (b) Common difference of the GP (r) = b a 1 n + 1 (c) Here, the first term (a) = 1 2 , the last term (b) = 16, number of geometric means = n ∴ Common ratio (r) = 16 1/2 1 n + 1 = (32) 1 n + 1 = 2 5 n + 1 ……… (i) By given, the third mean (g3) = 4 or, ar3 = 4 or, 1 2 . 2 5 n + 1 3 = 4 or, 2 15 n + 1 = 8 or, 2 15 n + 1 = 23 ∴ 15 n + 1 = 3 or, 3n + 3 =15 or n = 12 3 = 4. ∴ The required number of means is 4. (d) From (i), r = 2 5 4 + 1 = 2 5 5 = 2. ∴ The required mean (g4) = ar4 = 1 2 × 24 = 8. Example-5 There are n-GMs between 3 and 384. The ratio of the first mean and the last mean is 1:32. (a) What is the common ratio (r) in term of n. (b) Find the values of n and r. (c) Find the fifth mean. Solution: (a) Here, the first term (a) = 3, the last term (b) = 384, No. of GM’s = n, Again, taking the second and third parts, we get 8 x = y 8 or, 8 1 = y 8 or, y = 64 ∴ x = 1 and y = 64.

Algebra 172 Allied The Leading Mathematics-10 Sequence and Series 173 ∴ Common ratio (r) = b a 1 n + 1 = 384 3 1 n + 1 = 128 1 n + 1 = 2 7 n + 1 (b) Now, the first mean : the last mean = 1:32 or, ar arn = 1 32 or, 1 r n – 1 = 1 32 or, 1 2 7 n + 1 n – 1 = 1 25 or, 2 7 (n – 1) n + 1 = 25 ∴ 7n – 7 n + 1 = 5 Example-6 4 GMs are inserted between 3 and p. If the fourth mean is 48. (a) Find the common ratio (r). (b) Find the other three means. (c) Find the value of p. Solution: (a) Given, the first term (a) = 3, the last term (b) = p, No. of geometric means (n) = 4. Now, the fourth mean (g4) = 48 or, ar4 = 48 or, 3 × r4 = 48 or, r4 = 16 = 24 or, r = 2. (b) First mean (g1) = ar = 3× 2= 6, Second mean (g2) = ar2 = 3× 22 =12, Third mean (g3) = ar3 = 3× 23 = 24. Hence, the required remaining geometric means are 6, 12 and 24. (c) Common ratio (r) = p 3 1 4 + 1 r = b a 1 n + 1 or, 2 = p 3 1 5 or, 32 = p 3 or, p = 96. Therefore, the value of p is 96. Example-7 There are two numbers 8 and 18. (a) Find their AM and GM. (b) Prove that: AM > GM Solution: (a) Here, the first term (a) = 8, the last term (b) = 18. Now, we have AM = a + b 2 = 8 + 18 2 = 26 2 = 13 and GM = ab = 8 × 18 = 144 = 12. Since 15 > 12, so AM > GM. Proved. Example-8 The AM and GM between two numbers are 10 and 6 respectively. Find the numbers. Solution: Let ‘a’ and ‘b’ be both positive or negative numbers. Then, or, 7n – 7 = 5n + 5 or, 2n =12 or, n = 6 ∴ r = 2 7 n + 1 = 2 7 6 + 1 = 2. (c) The required fifth mean (g5) = ar5 = 3 × 25 = 96.

Algebra 172 Allied The Leading Mathematics-10 Sequence and Series 173 AM = 10 or, a + b 2 = 10 ⇒ a + b = 20 ….…. (i) and GM = 6 or, ab = 6 ⇒ ab = 36 ……….. (ii) Now, we have a – b = (a + b)2 – 4ab or, a – b = (20)2 – 4 × 36 or, a – b = 400 – 144 or, a – b = 256 or, a – b = 16 …….. (iii) Adding eqn (i) and eqn (iii), we have 2a = 36 ⇒ a = 18. Putting the value of a in eqn (i), we have 18 + b = 20 ⇒ b = 2. Hence, the required two numbers are 18 and 2 respectively. Example-9 The sum of three numbers in an arithmetic sequence is 45. If 1, 3 and 9 are added to them respectively, then they form a geometric sequence. Find the numbers. Solution: Let a – d, a, a + d be three numbers in an AS, then a – d + a + a + d = 15 or, 3a = 15 ⇒ a = 5. If 1, 3 and 9 are added to them respectively then the three numbers a – d + 1, a + 3, a + d + 9 are in GS, so, a + 3 a – d + 1 = a + d + 9 a + 3 or, 5 + 3 5 – d + 1 = 5 + d + 9 5 + 3 [ ∴ a = 5] or, 8 6 – d = 14 + d 8 or, 64 = 84 – 8d – d2 or, d2 + 8d – 20 = 0 or, d2 + 10d – 2d – 20 = 0 or, d(d + 10) – 2(d + 10) = 0 or, (d + 10)(d – 2) = 0 ∴ d = –10 or, 2 If a = 5 and d = –10 then the required three numbers are 5 + 10, 5, 5 – 10 = 15, 5, –5. If a = 5 and d = 2 then the required three numbers are 5 – 2, 5, 5 + 2 = 3, 5, 7. Note: If we take a - b = - 16, then a = 2 and b= 18. Both cases (a = 18, b = 2, and a = 2 and b = 18) have the same AM and GM.

Algebra 174 Allied The Leading Mathematics-10 Sequence and Series 175 Example-10 If the geometric mean between a and b is an + 1 + bn + 1 an + bn , find the value of n. Solution: Here, the geometric mean between a and b (GM) = an + 1 + bn + 1 an + bn . or, ab = an + 1 + bn + 1 an + bn or, a1/2b1/2 = an + 1 + bn + 1 an + bn or, an + 1/2b1/2 + a1/2bn + 1/2 = an + 1 + bn + 1 or, 0 = an + 1 – a1/2bn + 1/2 – an + 1/2b1/2 + bn + 1 or, a1/2(an + 1/2 – bn + 1/2) – b1/2(an + 1/2 – bn + 1/2) = 0 or, (an + 1/2 – bn + 1/2)(a1/2 – b1/2) = 0 or, an + 1/2 – bn + 1/2 = 0 [∴ a ≠ b => a1/2 ≠ b1/2] or, an + 1/2 = bn + 1/2 or, an + 1/2 bn + 1/2 = 1 or, ( a b)n + 1/2 = ( a b)0 ∴ n + 1 2 = 0 or, n = – 1 2. PRACTICE 6.3 Read Think Understand Do Keeping Skill Sharp 1. (a) What is geometric mean ? Define it. (b) What is the geometric mean between m and n ? (c) There are n-geometric means between a and b. How many terms does it have ? (d) What is the common ratio of the sequence having the first term a and the last term b where there are n-geometric means inserted between them? (e) Write the geometric mean between two terms 4 and 9. (f) What is the second term of arithmetic sequence, whose first and third terms are 105 and 53? (g) Write the relation between arithmetic mean (AM) and geometric mean (GM). (h) In which condition AM and GM are equal ? 2. Circle ( ) the correct answer. (a) If r is the geometric mean between p and q of an GP, which is the correct relation between p, q and r ? (i) r 2 = pq (ii) 2r = pq (iii) p2 = qr (iv) 2q = pr

Algebra 174 Allied The Leading Mathematics-10 Sequence and Series 175 (b) Which is the geometric mean between 1 4 and 1 9 ? (i) 1 3 (ii) 1 4 (iii) 1 5 (iv) 1 6 (c) If there are 3 geometric means between 2 and 162, which is the correct common ratio ? (i) 2 (ii) 3 (iii) 4 (iv) 5 (d) If the common ratio of a GP is 2 having the first term 2 and the last term 256, the 6th mean is .... . (i) 128 (ii) 64 (iii) 256 (iv) 512 (e) If n-geometric means is inserted between a and b then the nth geometric mean will be .... . (i) a b a n n + 1 (ii) a b a n n - 1 (iii) a a b n n + 1 (iv) a a b n n - 1 Check Your Performance 3. (i) (a) Define geometric mean (GM) between two numbers. (b) Find the GM between 7 and 28. (c) What is its common ratio ? Find it. (ii) If the 4th term and 6th term of a GP are 4 5 and 5 4 respectively then. (a) Find its 5th term. (b) Find its 10th term. (iii) If the 9th and 11th terms of a GP are 2x2 and 4x2 respectively, find the 5th term. (a) Find its common ratio and the first term. (b) Find its 5th term. 4. (i) p, 6, q, 216 are in GP. (a) What is q in the GP ? (b) Find the value of p. (c) Write the relation between p and q. (ii) If the geometric mean between x and 60 is 30 then find the value of x. 5. (i) 4, x – 6 and x + 9 are the first three terms of GP. (a) What is the value of x ? Find it. (b) Find the 5th term. (ii) The first three terms of a GP are k – 1, 2k + 1 and 7k – 1. Find the values of k and its 6th term in each case. (a) What is the value of k? Find it. (b) Find the 6th term. 6. (i) The numbers 1 32 and 32 are the first and the last terms of a GP. (a) If the GP has n-geometric means, what is the common ratio in term of n? (b) Insert 4 GM’s between these two numbers. (ii) 1 4, x, y, z , 64 are in GP. (a) What is its common ratio ? (b) Find the values of x, y and z. 7. (i) The first and the last numbers of a GP are 5 and 160 respectively. (a) If its third geometric mean is 40, how many terms are there in the GP ? (b) What is the common ratio of the GP ? Find it. (c) Find its other means. (ii) There are n-geometric means between 3 64 and 192. (a) If the second mean is 3, find the value of n. (b) What is the common ratio of the GP ? Find it. (c) Find its other means.

Algebra 176 Allied The Leading Mathematics-10 Sequence and Series 177 8. (i) When the second mean and fifth mean between 1 2 and 64 are in the ratio of 1:8 of a GP. (a) What is the common ratio (r) in term of n. (b) Find the values of n and r. (c) Find the fourth mean. (ii) Some geometric means are inserted between 11 25 and 1375. (a) Find the number of means when the ratio of the first mean to the last mean is 1:125. (b) What is the common ratio (r). (c) Find the geometric means. 9. (i) There are 3 geometric means between a and b. The first and the third means are 4 and 16 respectively. (a) Find the values of a and b. (b) Find the common ratio (r) of the GP. (ii) 7 geometric means are inserted between p and q such that the second and sixth means are 12 and 192 respectively. (a) Find the values of p and q. (b) Find the common ratio (r) of the GP. 10. For the two numbers; (i) 4 and 64 (ii) 3 and 27 (a) Find their AM and GM. (b) Verify that: AM > GM 11. (a) If the AM and GM between two numbers are 65 and 25 respectively, find the numbers. (b) If AM and GM between two numbers are 41 4 and 2 respectively, find them. 12. (a) The sum of three consecutive numbers in an AP is 24. If 2, 1 and 12 are added to them respectively then they form a GP. Find the numbers. (b) The product of three numbers in a GP is 8. If 1 is subtracted from the first number, the numbers form an AP. Find the numbers. 13. (a) If l, m, n are in AP and p, q, r are in GP, prove that: pm – n. qn – l . rl – m = 1. (b) If the arithmetic mean between a and b is an + 1 + bn + 1 an + bn , find the value of n. 3. (i) (b) 14 (c) 41/3 (ii) (a) 1 (b) 256 625 (iii) (a) 2 , x2 8 (b) x2 2 4. (i) (a) 36 (b) 1 (c) q = 36p (ii) 15 5. (i) (a) 0 or 16 (b) 81 4 or 625 4 (ii) (a) 0 or 4 (b) 1 or 729 6. (i) (a) 4 (b) 1 8 , 1 2 , 2 , 8 (ii) (a) 4 (b) 1, 4, 16 7. (i) (a) 4 (b) 2 (c) 10, 20, 40, 80 (ii) (a) 3 (b) 8 (c) 3 8 , 3, 24 8. (i) (a) 27/n + 1 (b) 6, 2 (c) 8 (ii) (a) 4 (b) 5 (c) 11 5 , 11, 55, 275 9. (i) (a) 2, 32 (b) 2 (ii) (a) 3, 384 (b) 2 10. (i) (a) 34, 16 (ii) 15, 9 11. (a) 125, 5 (b) 8, 1 2 12. (a) 1, 8, 15 or 25, 8, – 9 (b) 1, 2, 4 or 4, 2, 1 13. (b) 1 Answers

Algebra 176 Allied The Leading Mathematics-10 Sequence and Series 177 6.4 Sum of Geometric Series At the end of this topic, the students will be able to: ¾ solve the verbal problems related to geometric series. Learning Objectives Let us consider an example, a man gives Re. 1 on the 1st day, Rs. 2 on the 2nd day, Rs, 4 on the third day and so on by multiplying by 2 for 10 days to his friend. How much does he give to his friend on these 10 days? It forms a geometric series as 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512. By using calculator or adding each term, we obtain the sum Rs. 1023. i.e., 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 = 1023. This is the sum of the first 10 terms of GP. How can we take the sum by using formula? Suppose the sum of the first seven terms, S10 = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 ..........……… (i) Common ratio (r) = 2. So, multiplying by 2 in the relation (i), we get 2S7 = 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 ……..… (ii) By subtracting (i) from (ii), we get (2 – 1)S7 = – 1 + 1024 => S7 = 1023 1 = 1023 Suppose, a = the first term, d = the common difference, n = the number of terms, l = the last term, then the sum of the first n terms is denoted by Sn. So, Sn = a + ar + ar2 + ……… + arn-2 + arn-1 …….…..….. (i) Multiplying on both sides by r, we get r.Sn = ar + ar2 + ar3 + ….. + arn-1 + arn ………………. (ii) Subtracting the equation (i) from (ii), we get (r – 1)Sn = – a + arn or, Sn = a(rn – 1) r – 1 , r > 1 or, Sn = a(1 – rn ) 1 – r , r < 1 Also, Sn = arn – a r – 1 = r.arn – 1 – a r – 1 = rl – a r – 1 , r > 1. i.e., Sn = rl – a r – 1 , r > 1 or, Sn = a – rl 1 – r , r < 1 Points to be Remembered 1. The sum or additional form of an geometric progression is called an geometric series. 2. The sum of the first n-terms of an geometric series is calculated by the formulas as given below. Sn = a(rn – 1) r – 1 , r > 1 or, Sn = a(1 – rn ) 1 – r , r < 1 OR Sn = arn – a r – 1 , r > 1 or, Sn = a – rl 1 – r , r < 1 How much will pay on 30 days in total? Alert to Arithmetic Sum (i) Geometric series is not only addition form, it would be in subtraction form. (ii) The mentioned above formulas of the sum of geometric series hold for finite geometric series.

Algebra 178 Allied The Leading Mathematics-10 Sequence and Series 179 Example-1 The geometric series is 1 8 + 1 2 + 2 + 8 + …….. to 10 terms. (a) Which formula is used to find the sum of the given GS ? (i) Sn = a(rn – 1) r – 1 (ii) Sn = rl – a r – 1 (iii) Sn = a – rl 1 – r (iv) Both (ii) or (iii) (b) Find the common ratio of the given series. (c) Find the sum of the given series. Solution: (a) Sn = a(rn – 1) r – 1 is used to find the sum of the given GS. (b) Common ratio of the given series (r) = 1/2 1/8 = 1 2 × 8 1 = 4. (c) In the given series, the first term (a) = 1 8 , No. f terms (n) = 10 Now, we have Sum of the first 10 terms of the series (S10) = a(rn – 1) r – 1 = 1/8(410 – 1) 4 – 1 = 1048575 8 × 3 = 3495255 8 = 43690.625. Example-2 2048 + 1024 + 512 + ……. + 1 is a series. (a) Is the given series a geometric series ? Give reason. (b) Find the sum of the given series. Solution: (a) Here, in the series 2048 + 1024 + 512 + ……. + 1, Ratios, r1 = 1024 2048 = 1 2, r2 = 512 1024 = 1 2. ∴ The given series is geometric series. (b) First term (a) = 2048, Last term (l) = 1, Common Ratio (r) = 1 2, Sum of the series (Sn) = ? Now, we have Sum of the series (Sn) = a – rl 1 – r = 2048 – 1 8 . 1 1 – 1 2 = 4096 – 1 8 2 – 1 2 = 4095. Example-3 The sum of a series in GP with the first and the last terms 7 and 1701 respectively, is 2548. (a) Find the common difference. (b) Find the sum of its first 10 terms. Solution: (a) Given, the first term (a) = 7, the last term (l) = 1701, Now, Sum of the series in G.P (Sn) = 2548 or, rl – a r – 1 = 2548 or, r × 1701 – 7 r – 1 = 2548 or, 2548r – 2548 = 1701 – 7

Algebra 178 Allied The Leading Mathematics-10 Sequence and Series 179 or, 2548r – 1701 r = 2548 – 7 or, 847 r = 2541 or, r = 2541 847 = 3. Hence, the required common ratio is 3. (b) Sum of the first 10 terms (Sn) = arn – a r – 1 = 7 × 310 – 7 3 – 1 = 413336 2 = 206668. Example-4 The sum of a series in GP with the first and the last terms 7 and 1701 respectively, is 2548. Find the number of terms in the series if its common ratio is 3. Solution: Given, the first term (a) = 7, the last term (l) = 1701, Common ratio (r) = 3 Sum of the series in G.P (Sn) = 2548, No. of terms (n) = ? Now, we know that Sn = arn – a r – 1 or, 2548 = 7(3n – 1) 3 – 1 or, 2548 = 7(3n – 1) 2 or, 2548 × 2 7 = 3n –1 or, 3n = 728 + 1 or, 3n = 729 or, 3n = 36 ∴ n = 6 Hence, the required number of terms is 6. Example-5 If the sixth term is 16 times the second term and the sum of the first seven terms is 127 4 in a geometric series. (a) Find the positive common ratio. (b) Find the first term of the series. Solution: (a) Let, a and r be the first term and common ratio of a geometric series respectively, then by question, Sixth term (t6) = 16 × Second term or, ar5 = 16 × ar or, r 4 = 24 ∴ r = 2 Example-6 8 + 16 + 32 + …………. + 2048 is a GP. . (a) Add the next three terms in the given series. (b) Find the sum of new GP. Solution: (a) In a GP, 8 + 16 + 32 + …………. + 2048, the first term (a) = 8, Common ratio (r) = 16 8 = 2, the last term (l) = 2048, No. of terms (n) = ? (b) Sum of the first seven terms (S7) = 127 4 or, a(27 – 1) 2 – 1 = 127 4 or, a(128 – 1) 2 – 1 = 127 4 or, a = 127 127 × 4 or, a = 1 4.

Algebra 180 Allied The Leading Mathematics-10 Sequence and Series 181 Now, we know that l = arn – 1 or, 2048 = 8 × 2n – 1 or, 2048 8 = 2n – 1 or, 256 = 2n – 1 or, 28 = 2n – 1 ∴ 8 = n – 1 => n = 9 (b) By question, adding next three terms, then total no. of terms (N) = 9 + 3 = 12 ∴ Sum of the first 12 terms of new GP (S12) = a(rn – 1) r – 1 = 8(212 – 1) 2 – 1 = 8 × 4095 = 32760. Example-7 A rich man saves Re. 1 for the first day, Rs. 2 for the second day, Rs. 4 for the third day and so on at a pingpong. How much money did he save at the pingpong after 30 days? Solution: Here, The saving amount for the first, the second, the third days are as a series 1 + 2 + 4 + ...., which is a geometric series, in which, Common ratio (r) = 2 1 = 4 2 = 2 First term (a) = 1, No. of days (n) = 30 Now, we have Sn = a(rn – 1) r – 1 Sum of the saving amount after 30 days (S30) = 1(230 – 1) r – 1 = 1073741824 – 1 1 = 1073741823. Thus, he saved Rs. 1073741823. PRACTICE 6.4 Read Think Understand Do Keeping Skill Sharp 1. (a) Write the definition of geometric series with example. (b) Write the formula to find the sum of a geometric sequence having the first term 'a', common ratio 'r' and number of terms 'n'. (c) Write the formula to find the sum of a geometric sequence having the first term 'a', common ratio 'r' and the last term 'l'. (d) What are the meanings of a and r in the formula of the sum of an geometric series Sn = a(rn – 1) r – 1 ?

Algebra 180 Allied The Leading Mathematics-10 Sequence and Series 181 2. Circle ( ) the correct answer. (a) What is the sum of the first n-terms of an geometric series with the first term a and common ratio r? (i) Sn = a(rn – 1) r – 1 (ii) Sn = rl – a r – 1 (iii) Sn = a(1 – rn ) 1 – r (iv) All of the above (b) What is the common ratio of the geometric series 2 + 4 + 8 + 16 + ..... ? (i) 1 (ii) 2 (iii) 4 (iv) 8 (c) Which is a geometric series ? (i) 4 + 6 + 8 + 10 + .... (ii) 4 – 6 + 8 – 10 + .... (iii) 27 + 9 + 3 + 1 + .... (iv) 5 + 7 + 10 + 14 + .... Check Your Performance 3. Find the sum of the following series: (i) 3 + 6 + 12 + 24 + ……. to 11 terms. (ii) 2 + 6 + 18 + 54 + ……. to 9 terms. (a) Find the common ratio of the given series. (b) Find the sum of the given series. 4. Find the sum of the following series: (i) 8 + 12 + 18 + 27 + ……. + 60.75. (ii) 31 + 32 + 33 + 34 + …………….. + 38 . (a) Is the given series a geometric series ? Give reason. (b) Find the sum of the given series. 5. (i) The sum of the first three terms of a GP with the first term 5 is 35. (a) Find the common ratio. (b) Find the sum of its first 10 terms. (ii) The sum of the first seven terms of a GS having the common ratio 2 is 1016. (a) Find the first term of the GS. (b) Find the sum of its first 12 terms. 6. (i) The sum of a geometric series is 63. If its first term and common ratio are 1 and 2 respectively, (a) How many terms are there in the series? (b) Find the sum of its first 10 terms. (ii) The first and the last terms of a geometric series having sum 484, are 4 and 324 respectively. (a) Find its common ratio. (b) Find its number of terms. (iii) The series 2 + 8 + 32 + ….… has the sum 682. (a) Find the common ratio of the series. (b) How many terms does it have ?

Algebra 182 Allied The Leading Mathematics-10 Sequence and Series PB 7. (i) The first term of the geometric series is 2 and the fourth term is 54. (a) Find the common ratio of the GS. (b) Find the sum of the first eight terms. (ii) The 2nd and the 4th terms of a GP having positive common ratio are – 6 and – 24 respectively. (a) Find the common ratio of the GS. (b) Find the sum of the first eight terms. (iii) The sum of the first four terms of a GS with positive common ratio is 15 and that of the first eight terms is 255. (a) Find the first term and common ratio of the GS. (c) Find the sum of its first ten terms. (iv) The sixth term is 16 times of the second term and the sum of the first seven terms is 127 4 in a geometric series having positive common ratio. (a) Find the common ratio and the first term of the GS. (b) Find the sum of the first nine terms. 8. (i) 8 + 16 + 32 + …………. + 2048 is a GP. (a) Add the next three terms in the given series. (b) Find the sum of new GP. (ii) A GP has the fourth term 200 and the fifth term 400. How many terms should be taken to make the sum 12775 of the GP? 9. (i) There are nine varieties of birds in a zoo. The number of each variety is double at the number of another variety. If the number in the first variety is 3, find the total number of birds in the zoo. (ii) A boy borrows Rs. 32766 without interest and doubles the loan in each installment being double of preceding one. If the amount of the first installment is Rs. 2, find the number of installments in which he pays to complete the loan. (iii) Manish borrows Rs. 4368 without interest which he promises to pay in 6 annual installments and each installment being triple of the preceding one. Find the first and the last installments. 3. (i) (a) 2 (b) 6141 (ii) (a) 2 (b) 19682 4. (i) (b)166.25 (ii) (b) 9840 5. (i) (a) – 3 or 2 (b) – 73810 or 5115 (ii) (a) 8 (b) 32760 6. (i) (a) 6 (b) 1023 (ii) (a) 3 (b) 5 7. (i) (a) 3 (b) 6560 (ii) (a) 2 (b) 765 (iii) (a) 1, 2 (b) 1023 (iv) (a) 2, 1 4 (b) 127 3 4 8. (i) 32760 (ii) 9 9. (i) 1533 (ii) 14 (iii) Rs. 12, Rs. 2916 Answers

PB Allied The Leading Mathematics-10 Quadratic Equation 183 Algebra 7.1 Solution of Quadratic Equation At the end of this topic, the students will be able to: ¾ solve the quadratic equations. Learning Objectives I Review on Quadratic Equation Activity 1 Anita and Bimi are two best friends of a school who study in class 10 in different sections A and B in equal areas 600 sq. ft. respectively. They discuss the sizes of their own classrooms. Anita says, "The ratio of the length and breadth of my classroom A is 3:2." Bimi says, "My classroom B has its breadth 1 feet less by its length." Now, what are the length and breadth of the classrooms of sections A and B ? Discuss. What is the formula to find the area of rectangle ? How can you express the above conversation in the mathematical statement ? Can you express the above relations in quadratic form? For Anita's classroom section A, length (l A) = 3x and breadth (bA) = 2x (suppose) Then, Area of the classroom section A = 600 or, 3x × 2x = 600 => x2 = 100 ∴ x = 10. How ? Can you do ? Hence, Length of classroom section A (l A) = 3 × 10 = 30 ft and Breadth (bA) = 2 × 10 = 20 ft. For Bimi's classroom section B, length (l B) = x and breadth (bB) = x – 1 (suppose) Then, Area of the classroom section B = 600 or, x(x – 1) = 600 => x2 – x – 600 = 0 By solving this quadratic equation, we get x = – 24 or 25. How ? Can you do ? Hence, Length of classroom section C (l B) = 25 ft and Breadth (bB) = 25 – 1 = 24 ft. Here, the equations x2 = 100 and x2 – x – 600 = 0 are in the second degrees. These types of equations in the second degree are called quadratic equations in a single variable x. x x x2 10 10 10 10 100 100 x – 10 x + 10 x – 10 x – 10 x x 10 x 10 – = = x2 – 100 x x x x x 1 1 1 x2 25 24 24 24 600 600 x + 24 x – 25 – – = CHAPTER 7 QUADRATIC EQUATION

184 Allied The Leading Mathematics-10 Quadratic Equation 185 Algebra The word Quadratic is derived from the word Quadratus which means square. The quadratic equation and its solution was used from 2000 BC by Babylonian, Egyptian, Arabian, Chinese and Indian mathematicians. But, they had no notion of 'equation'. What they developed was an algorithmic approach for solving problems which, in our terminology, would give rise to a quadratic equation. After a long time, at the end of the 16th Century, mathematical notation and symbolism were introduced by a French amateur-mathematician François Viète (1540-1603). In 1637, a French mathematician René Descartes (1596-1650) published the quadratic formula has adopted the form we know today. There are many scenarios where a quadratic equation is actually used in everyday life, as when calculating areas, different gaming, entertainments, determining a product's profit or formulating the speed of an object. It is used in many areas of mathematics, science and technology, and economic sectors. II Roots of Quadratic Equations The value or values of the variable x that make the equation true are called the root/s of the equation. Consider a quadratic equation x2 + x − 6 = 0. Suppose, x2 + x − 6 = 0 = y The, we obtain two equation as y = 0 ......... (i) , which is the equation of x-axis. This is well known straight line. and y = x2 + x − 6 .......... (ii), which is also a quadratic equation. Make table for drawing the graph, x 0 1 − 1 2 − 2 3 − 3 − 4 y − 6 − 4 − 6 0 − 4 6 0 6 Draw graph on the basis of the above table, we obtain the U-shaped curve from the quadratic equation (ii) as shown in the above graph. This types of curve is called Parabola. This parabola cuts the straight line obtained by the equation (i) i.e. x-axis at two points with x-intercepts 2 and − 3. These two intercepts are called the roots of the quadratic equation x2 + x − 6 = 0. A study of quadratic equation in one variable begins with one fundamental theorem, known as the Fundamental Theorem of Algebra (FTA) and Zero Factor Theorem (ZFT). The FTA states that every quadratic equation has at least one root. We assume this theorem without proof. To find the François Viète Quadratic equation is used in bridge. Cooking Parabolic shaped disc −3 2 Satellite Parabolic shaped disc

184 Allied The Leading Mathematics-10 Quadratic Equation 185 Algebra root or roots of a quadratic equation means to solve the quadratic equation. The ZFT sates that Given any two (real) numbers a and b, if ab = 0, then either a = 0 or b = 0 or both a = 0 and b = 0. III Solution of Quadratic Equations A complete quadratic equation of the form ax2 + bx + c = 0, a ≠ 0 may be solved in various ways. We discuss the following three different methods: A. Method of factorization B. Method of completing the squares C. Method of using quadratic formulae A Solutions of Quadratic Equations by Factorization This method may be carried out in the following four steps: 1. Express the given equation in the standard form by transposing every term to the left side before or after simplification; 2. Factorize the left member; 3. Put each factor = 0; 4. Solve each of the resulting linear equations. For example, consider a quadratic equation x2 + 5x + 6 = 0 The equation is complete and is in the standard form. Factorizing the left member, we get x2 + (2 + 3)x + 6 = x2 + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2) (x + 3) The given equation can therefore be written as (x + 2) (x + 3) = 0 Applying the zero factor theorem, if ab = 0, then either a = 0 and/or b = 0, we get Either, x + 2 = 0 OR, x + 3 = 0 ∴ x = – 2 or, x = – 3. That is, the required roots are x = – 2 or, x = – 3. "Alternatively" x2 + 5x + 6 can also be factorized by using tiles as alongside. B Solutions of Quadratic Equations by Completing Square Solving quadratic equations by factorization is not so easy in all cases. Completing the squares of both sides often makes it easier to solve a quadratic equation. This can be done in two ways. We illustrate them with concrete example. Consider a quadratic equation x2 + 2x – 3 = 0. Prime factors of 6 are 2 and 3. Constant term is positive. So, 5 = 2 + 3 x2 – (sum of roots)x + Product of roots x x x2 + + = 5 6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x x x x x x + 3 x + 2 5x

186 Allied The Leading Mathematics-10 Quadratic Equation 187 Algebra Here, the coefficient of x2 is 1. Add and subtract (half the coefficient x)2 to the left side. Then, x2 + 2x + 12 – 12 – 3 = 0, i.e., (x + 1)2 – 4 = 0 i.e., (x + 1)2 = 4 = 22 Taking the square root of each side, we get x + 1 = + 2 That is, x + 1 = 2 or x + 1 = –2 ∴ x = 1 or x = –3. C Solutions of Quadratic Equations by Using Quadratic Formulae The general (or standard) form of a quadratic equation is, ax2 + bx + c = 0 . (a ≠ 0) All equations considered so far can be obtained from this equation by giving particular values to the constants a, b and c. We first derive a formula that gives the roots of this equation directly. For this we follow the method of completing squares. Rewriting the given equation ax2 + bx + c = 0 ..................... (i) in the form ax2 + bx = – c Dividing throughout by a (the coefficient of x2 ), to find x2 + b a x = – c a or, x2 + 2 b 2ax = – c a ..................... (ii) Adding b 2a 2 on both sides, to find x2 + 2 b 2ax + b 2a 2 = – c a + b 2a 2 i.e., x + b 2a 2 = – c a + b2 4a2 = b2 – 4ac 4a2 Taking the square roots of both sides, we get x + b 2a = ± b2 – 4ac 2a Subtracting – b 2a from each side, we get x = – b 2a ± b2 – 4ac 2a = – b ± b2 – 4ac 2a . Note : The procedure followed here is exactly the same as the one we used in the method of completing the squares. Instead of going through the various steps, we could use the formulae directly, hence the name for the rule. For example, Consider a quadratic equation 3x2 – 5x + 2 = 0. x x x2 + – = 2 3 1 1 x 2x x x 1 1 x + 1 x + 1 – 4 1 1 1 1 1 Alternative way to factorise x2 + 2x – 3 for completing square Remarks: In quadratic or root formula, a. if b2 – 4ac = D (say), D = 0, its roots are real and equal. b. if D > 0 and perfect square, its roots are real, unequal and rational. c. if D > 0 and not perfect square, its roots are real, unequal and irrational. d. if D < 0, its roots are unequal and imaginary.

186 Allied The Leading Mathematics-10 Quadratic Equation 187 Algebra In the equation 3x2 – 5x + 2 = 0, a = 3, b = –5, c = 2. Substituting these values in the quadratic formulae x = – b ± b2 – 4ac 2a , we get x = 5 ± (– 5)2 – 4 (3) (2) 2(3) = 5 ± 1 6 = 5 ± 1 6 = 6 6 or 4 6 = 1 or 2 3. So, the required roots are x = 1 or 2 3. Example-1 (a) How many roots are there in the quadratic equation x2 – 5x + 6 = 0 ? (b) Factorize the left expression and solve it. (c) Check the answers. Solution: (a) There are two roots in the quadratic equation x2 – 5x + 6. (b) Here, x2 – 5x + 6 = 0 or, x2 – (2 + 3)x + 6 = 0 or, x2 – 2x – 3x + 6 = 0 or, x(x – 2) – 3(x – 2) = 0 or, (x – 2) (x – 3) = 0 Either, x – 2 = 0 => x = 2. OR, x – 3 = 0 => x = 3. (c) For x = 2, x2 – 5x + 6 = 0 or, 22 – 5 × 2 + 6 = 0 or, 4 – 10 + 6 = 0 or, 0 = 0, which is true. Hence, the required solution is x = 2 or 3. Example-2 Solve by factorization method: (a) 2x2 + 5x – 3 = 0 (b) 4x2 – 12x – 7 = 0 Solution: (a) Here, 2x2 + 5x – 3 = 0 or, 2x2 + (6 – 1)x – 3 = 0 or, 2x2 + 6x – x – 3 = 0 or, 2x(x + 3) – 1(x + 3) = 0 or, (x + 3) (2x – 1) = 0. Either, x + 3 = 0 => x = – 3 OR, 2x – 1 = 0 => x = 1 2 Hence, x = 1 2, – 3 Factors of 6 = 2 × 3 or 1 × 6 Constant is positive (+). So, we should need 5 by adding two factors of 6. Here, 2 + 3 = 5 For x = 3, x2 – 5x + 6 = 0 or, 32 – 5 × 3 + 6 = 0 or, 9 – 15 + 6 = 0 or, 0 = 0, which is true. Product of 2 and 3 = 6 Constant is negative (–). So, we should need 5 by subtracting two factors of 6. Factors of 6 = 2 × 3 or 1 × 6 Here, 3 – 2 ≠ 5 or 2 – 3 ≠ 5 But, 6 – 1 = 5

188 Allied The Leading Mathematics-10 Quadratic Equation 189 Algebra (b) Here, 4x2 – 12x – 7 = 0 or, 4x2 – (14 – 2)x – 7 = 0 or, 4x2 – 14x + 2x – 7 = 0 or, 2x(2x – 7) + 1(2x – 7) = 0 or, (2x – 7) (2x + 1) = 0 Hence, either 2x – 7 = 0 ∴ x = 7 2 . OR, 2x + 1 = 0 ∴ x = – 1 2 . Example-3 Solve: (a) 2 x2 – 1 – 1 x – 1 = 1 (b) Does x = 1 hold for the given equation? Why? Solution: Here, 2 x2 – 1 – 1 x – 1 = 1 or, 2 (x + 1) (x – 1) – 1 x – 1 = 1 or, 2 – (x + 1) (x + 1) (x – 1) = 1 or, 2 – x – 1 x2 – 1 = 1 or, x2 – 1 = 1 – x or, x2 + x – 2 = 0 [ +1 = +2 – 1 and (+2)×(–1)= –2] or, x2 + 2x – x – 2 = 0 or, x(x + 2) – 1(x + 2) = 0 or, (x – 1) (x + 2) = 0 Either, x – 1 = 0 ∴ x = 1 OR, x + 2 = 0 ∴ x = – 2 Example-4 Solve the quadratic equation x – 2 x + 2 – x + 2 x – 2 = 2 1 2 and check its solutions. Solution: Here, x – 2 x + 2 + x + 2 x – 2 = 21 2 or, (x – 2)2 + (x + 2)2 (x + 2) (x – 2) = 5 2 or, x2 – 4x + 4 + x2 + 4x + 4 x2 – 22 = 5 2 or, 2x2 + 8 x2 – 4 = 5 2 or, 4x2 + 16 = 5x2 – 20 or, 5x2 – 4x2 = 16 + 20 or, x2 = 36 ∴ x = ± 6 Product of 4 and 7 = 28 Constant is negative (–). So, we should need 12 by subtracting two factors of 28. Factors of 28 = 2 × 2 × 7 = 4 × 7 or 2 × 14 Here, 7 – 4 ≠ 12 But, 14 – 2 = 12 Put x = 1 in the given equation, 2 12 – 1 – 1 1 – 1 = 1 or, 2 0 – 1 0 = 1 or, ∞ – ∞ = 1 or, ∞ = 1, which is not defined. So, x = 1 does not hold for the given equation. When the denominators of fractions are different, we use "butterfly multiplication". a b + d c "Alternatively for x2 = 36" x2 = 36 or, x2 – 36 = 0 or, x2 – 62 = 0 or, (x + 6)(x – 6) = 0 Either, x + 6 = 0 ∴ x = – 6 Or, x – 6 = 0 ∴ x = 6

188 Allied The Leading Mathematics-10 Quadratic Equation 189 Algebra For checking; For x = 6, x – 2 x + 2 + x + 2 x – 2 = 21 2 or, 6 – 2 6 + 2 + 6 + 2 6 – 2 = 5 2 or, 4 8 + 8 4 = 5 2 or, 1 2 + 2 = 5 2 or, 5 2 = 5 2, which is true. Hence, the answer x = ± 6 is correct. Example-5 Solve by the method of completing the squares: 3x2 + 2x – 1 = 0 Solution: Here, 3x2 + 2x – 1 = 0 3x2 + 2x = 1 or, x2 + 2 3 x = 1 3 [ Dividing on both sides by 3] or, x2 + 2 3x + 1 3 2 = 1 3 + 1 3 2 [ Adding 1 3 2 on both sides] or, x + 1 3 2 = 1 3 + 1 9 or, x + 1 3 2 = 4 9. i.e., x + 1 3 2 = 2 3 2 Taking the square root of each side, we have x + 1 3 = + 2 3 That is, x + 1 3 = 2 3 or – 2 3 => x = 2 3 – 1 3 = 1 3 or x = – 2 3 – 1 3 = – 2 – 1 3 = – 3 3 = –1. "Alternatively" Here, 3x2 + 2x – 1 = 0 or, 3x2 + 2x = 1 or, 32 x2 + 2x.3 = 3 [ Multiplying on both sides by 3] or, (3x)2 + 2(3x).1 + 12 = 3 + 1 [ Adding 1 on both sides] or, (3x + 1)2 = 4 or, (3x + 1)2 = 22 ∴ 3x + 1 = ± 2 For x = – 6, x – 2 x + 2 + x + 2 x – 2 = 21 2 or, – 6 – 2 – 6 + 2 + – 6 + 2 – 6 – 2 = 5 2 or, – 8 – 4 + – 4 – 8 = 5 2 or, 2 + 1 2 = 5 2 or, 5 2 = 5 2, which is true.

190 Allied The Leading Mathematics-10 Quadratic Equation 191 Algebra Taking + sign, 3x = 2 – 1 = 1 => x = 1 3 Taking – sign, 3x = – 2 – 1 = – 3 => x = – 3 3 = –1 Hence, x = 1 3 or – 1. Example-6 Solve the following equations by using root formulae: 3x2 – x – 1 = 0 Solution: Comparing the given equation 3x2 – x – 1 = 0 with ax2 + bx + c = 0, we have a = 3, b = –1, c = –1. Substituting in the formulae x = – b ± b2 – 4ac 2a , we get x = 1 ± (– 1)2 – 4(3) (– 1) 2(3) = 1 ± 13 6 = 1 + 13 6 or 1 – 13 6 Example-7 Using root formulae or otherwise solve the equation: x – 4 4 + 6 x + 1 = 5 4 Solution: Here, x – 4 4 + 6 x + 1 = 5 4 or, x – 4 4 – 5 4 = – 6 x + 1 So, x – 4 – 5 4 = – 6 x + 1 or, (x + 1) (x – 9) = – 24 Further simplification yields x2 – 8x + 15 = 0 (a) Solution by factorization method: Here, x2 – 8x + 15 = 0, i.e., (x – 3) (x – 5) = 0. Hence, x = 3 and x = 5. (b) Solution by using root formulae: Here, a = 1, b = – 8 and c = 15. Substituting in the formulae x = – b ± b2 – 4ac 2a , we get x = – (– 8) ± (–8)2 – 4 (1).(15) 2(1) = 8 ± 64 – 60 2 = 8 ± 2 2 = 3 or 5.

190 Allied The Leading Mathematics-10 Quadratic Equation 191 Algebra PRACTICE 7.1 Read Think Understand Do Keeping Skill Sharp 1. (a) What is quadratic equation ? Define it with example. (b) What is factorization? (c) If x(x + 1) = 0, what are the values of x? (d) If (x – 2)(x + 1) = 0, write down the values of x. (e) If 0.09x2 = 1, then x = ...... . (f) What are the root of the quadratic equation x2 + px + q = 0? 2. (a) The degree of quadratic equation is .......... . (i) 0 (ii) 1 (iii) 2 (iv) 3 (b) Which is standard form of a quadratic equation ? (i) x2 + bx + c = 0 (ii) ax2 + bx + c = 0 (iii) ac2 + x + c = 0 (iv) px2 + qx + 1 = 0 (c) Which is quadratic equation among the following equations ? (i) 2x = 3 – 3x3 (ii) 2x + 1 = 0 (iii) 2x(x + 4) = 2x2 + 3 (iv) 2x – 1 = 3x(x + 2) (d) The roots of quadratic equation px2 + qx + r = 0 is .......... . (i) – q ± q2 – 4pr 2p (ii) – b ± b2 – 4ac 2a (iii) – p ± q2 – 4pq 2r (iv) – b + b2 – 4ac 2a (e) Which are the roots of mx2 – n = 0 ? (i) + m n (ii) + m n (iii) + n m (iv) + n m Check Your Performance 3. (a) How many roots are there in the quadratic equation x2 + 5x + 6 = 0 ? (b) Factorize the left expression and solve it. (c) Check the answers. 4. (a) Write the roots of kx2 + mx – n = 0. (b) Solve and check: x2 + 3x – 18 = 0. 5. Solve the following equations by factorization and check the solutions. (a) x2 + 4x + 3 = 0 (b) x2 – 9x + 18 = 0 (c) 3x2 – 14x + 8 = 0 (d) 5x2 + 6x – 8 = 0 (e) x2 5 – x 3 – 8 15 = 0 (f) x2 + (a + b)x + ab = 0 (g) 3 + 2 x – 1 x2 = 0 (h) 4 y2 – 1 y = 0 (i) 2x2 + x – 3 = 0

192 Allied The Leading Mathematics-10 Quadratic Equation 193 Algebra 6. Solve the following equations by factorization and check the solutions. (a) x – 1 x – 6 = 3x – 11 4x – 21 (b) x 1 + x + x + 1 x = 21 6 (c) x + 3 x – 3 + x – 3 x + 3 = 5 2 (d) 4 x – 1 – 5 x + 2 = 3 x (e) x + 3 x + 2 – 2x – 3 x – 1 = x – 3 2 – x (f) 3 x – 4 x – 1 + 5 x + 2 = 0 7. Solve and testM (a) (x – 4) (x + 6) + (x + 2) (x – 2) = 56 (b) (3x – 2) (2x + 3) + 2 = 0 8. Solve each of the following equations by completing the square and test. (a) x2 + 4x + 3 = 0 (b) x2 + 6x – 7 = 0 (c) x2 – 6x – 7 = 0 9. Use the method of completing squares to solve the following equations. (a) 6x2 – 7x + 1 = 0 (b) 5u2 – 3u – 2 = 0 (c) 6x2 + 11x – 35 = 0 (d) − 5x2 = 9 + 46x (e) (w – 6)(w – 6) = 81 (f) 4 + 11 3 x – 5x2 = 0 10. Solve the following equations by using root formulae and test. (a) x2 – 4 = 0 (b) m2 = 7m (c) 3x2 + 4x + 1 = 0 (d) 3u2 – 4u + 1 = 0 (e) 4y2 – 3y – 5 = 0 (f) 4z2 + 2z – 5 = 0 11. Solve the following equations by using root formulae. (a) 3x – 7 2x – 5 = x + 1 x – 1 (b) 20 + x2 15 = 11 – 3x2 + 5 10 (c) x2 (a – b) – x(a + b)+2b= 0 3. (a) Two roots (b) (x + 2) (x + 3), – 2, – 3 4. (a) – m + m2 + 4kn 2k or, – m – m2 + 4kn 2k (b) – 6 or 3 5. (a) –1, –3 (b) 3, 6 (c) 4, 2 3 (d) 4 5 , – 2 (e) 8 3, – 1 (f) – a, – b (g) –1, 1 3 (h) – 1, 4 5 (i) –3 2, 1 6. (a) – 9 or 5 (b) – 3 or 2 (c) ± 9 (d) – 1 2 , 3 (e) 0, 4 (f) – 1 2 , 3 7. (a) – 7 or 6 (b) – 4 3 or 1 2 8.(a) −1, – 3 (b) –7 or 1 (c) 7 or – 1 9. (a) 1 or 1 6 (b) 1 or – 2 5 (c) 5 3 or − 7 2 (d) –9 or − 1 5 (e) 15 or –3 (f) 4 3 or – 3 5 10. (a) ± 2 (b) 7 or 0 (c) – 1 3 or –1 (d) 1 or 1 3 (e) 3 + 89 8 or 3 – 89 8 (f) –1 + 21 4 or –1 – 21 4 11. (a) 4 or 3 (b) ± 5 (c) 1 or 2b a – b Answers Project Work Rasmita's classroom has the area of 600 sq. ft. She says, "The breadth of my classroom is less by 1 6 th of its length in 20 feet." What is the formula to find the area of rectangle ? How can you express the above conversation in the mathematical statement ? Can you express the above relations in quadratic form? What are the length and breadth of the classroom ? Calculate.

192 Allied The Leading Mathematics-10 Quadratic Equation 193 Algebra 7.2 Verbal Problems on Quadratic Equations At the end of this topic, the students will be able to: ¾ solve the verbal (word) problems in the case of quadratic equations. Learning Objectives I Introduction Activity 1 Two brothers of the ages 15 and 18 years are sitting in the square garden of the area 25 sq. meter. They are thinking how much lengths of the garden are they sitting in and how many years the product of their age will be 460. But, they are confusing to solve it. How can we solve the problem ? Discuss. In the case of solving this problem, we face different types of quadratic equations as below. A quadratic equation of second degree of the form ax2 + bx + c = 0; a, b, c ∈ R and a ≠ 0 may be solved in various ways. There are at least two roots in the quadratic equation. If the quadratic equation is perfect, it has only one root. The roots of the quadratic equations are as follows: SN Quadratic equation Roots Examples 1. Pure quadratic equation ax2 = c (a ≠ 0) x = + c a 2x2 – 8 = 0 => x2 = 4 ∴ x = + 4 = + 2 2. Affected quadratic equation ax2 + bx = 0 (a ≠ 0) x = 0, – b a 3x2 + 2x = 0 => x(3x + 2) = 0 ∴ x = 0, – 2 3 3. Perfect quadratic equation x2 + 2abx + b2 = 0 (a, b ≠ 0) x = b, b x2 + 6x + 9 = 0 => (x + 3)2 = 0 => (x + 3)(x + 3) = 0 ∴ x = 3, 3 4. Complete quadratic equation ax2 + bx + c = 0 (a, b, c ≠ 0) x = – b ± b2 – 4ac 2a ; b2 – 4ac = D < 0, x2 – 5x + 6 = 0 => x2 – 2x – 3x + 6 = 0 or, (x – 2)(x – 3) = 0 ∴ x = 2, 3 Note: In quadratic or root formula, a. if b2 – 4ac = 0, its roots are real and equal. b. if b2 – 4ac > 0 and perfect square, its roots are real, unequal and rational. c. if b2 – 4ac > 0 and not perfect square, its roots are real, unequal and irrational. d. if b2 – 4ac < 0, its roots are unequal and imaginary.

194 Allied The Leading Mathematics-10 Quadratic Equation 195 Algebra We discuss the following three different methods that are mentioned in previous lesson: (a) Method of factorization (b) Method of completing the squares (c) Method of using quadratic formulae Many mathematical and life problems give rise to various kinds of quadratic equations. We know the quadratic equation is known to have two solutions; but it may so happen that one of the two solutions may be unacceptable in physical and life situation. For instance, a negative solution is usually unacceptable in many life related problems. Sometimes, fraction or irrational or nonreal solutions also become unacceptable. It is therefore always advisable to see whether a given solution is acceptable or not. If the root of the equation does not satisfy for the all conditions, it will be rejected. Sometimes the process of squaring both sides of the equations yields unacceptable solution. We follow some steps of solving verbal/word problems on Quadratic equations: Step I: Understand the problem and draw a picture if necessary. Step II: Denote the unknown quantities by x, y, z, u, v, etc. Step III: Use the conditions of the problem to establish in unknown quantities. Step IV: Use the equations to establish one quadratic equation in one unknown variable. Step V: Solve the equation using any one method to find the value of the unknown variable and reject unsuitable number for the given condition. Step VI: Check the work and test the answer for the given condition. We shall discuss the applications of quadratic equations in various sectors. II Applications of Quadratic Equations on Numbers We have solved number applications that involved comparison on numbers, consecutive numbers, even and odd integers, reciprocal numbers, two-digit numbers, etc. in quadratic equations. The simple information and notation of the mentioned problems on numbers are as below. SN Model of Numbers Definition Notations Examples 1 Successive Numbers (Sequence) Numbers that follow each other continuously in the order to ascending or descending under certain rule i) x, x + d, x + 2d, .... ii) x, 2x, 3x, .... iii) x, x2 , x3 , .... (x ≠ 1, 0) iv) x, xr, xr2 , .... 1, 3, 5, ... 2, 4, 6, ... 2, 4, 8, ... 3, 6, 12, ... 2. Consecutive Numbers Numbers that follow each other continuously in the order from smallest to largest x, x + 1, x + 2, x + 3, .... 2, 3, 4, ... 3. Consecutive odd Numbers Successive numbers increasing by 2 started from odd number (x) x, x + 2, x + 4, x + 6, .... ..., x – 2, x, x + 2, ... 3, 5, 7, ...

194 Allied The Leading Mathematics-10 Quadratic Equation 195 Algebra 4. Consecutive even Numbers Successive numbers increasing by 2 started from even number (x) x, x + 2, x + 4, x + 6, .... ..., x – 2, x, x + 2, ... 4, 6, 8, ... 5. Reciprocal Numbers Inverse (multiplicative) of a number i.e, a number and dividing 1 by the number or turning upside-down x and 1 x such that x × 1 x = 1 2 and 1 2 6. Two-digit number and Its reverse A number that has two digits and it starts from the number 10 and end on the number 99. 10x + y, x in tens, y in ones and its inverse 10y + x, y in tens, x in ones 19 and 91 Example-1 The sum of two numbers is 9 and their product is 18. (a) Express the given problem in the mathematical notations. (b) Convert them into a single variable equation. Name the type of the equation. (c) Solve this equation and test in the given conditions. (d) When 1 is added to each number, what percentage will the new product be increased than the original product ? Solution: (a) Let two numbers be x and y. Then, by question, x + y = 9 => y = 9 – x .......... (i) Again, xy = 18 .......... (ii) (b) Converting the eqn (i) and (ii) in a single variable, we get x(9 – x) = 18 or, x2 – 9x + 18 = 0, which is quadratic equation. (c) x2 – 9x + 18 = 0 or, x2 – (6 + 3)x + 18 = 0 or, x2 – 6x – 3x + 18 = 0 or, x(x – 6) – 3(x – 6) = 0 or, (x – 6)(x – 3) = 0 Either, x – 6 = 0 ∴ x = 6. Or, x – 3 = 0 ∴ x = 3. If x = 6, from eqn (i), y = 9 – 6 = 3. If x = 3, from eqn (i), y = 9 – 3 = 6. For testing, the sun of 6 and 3 or 3 or 6 is 18 and their product is 18. Hence, the required two numbers are 6 and 3 or 3 and 6. (d) When 1 is added to each number, the two new numbers will be 3+1 = 4 and 6+1 = 7. The product of these two new numbers is 4 × 7 = 28. ∴ Increased percentage in 28 than 19 = 28 – 19 19 × 100% = 9 19 × 100% = 47.37% "Alternatively for two numbers" (a) Let one of the numbers be x. Then the other number will be 9 – x, since their sum is 9. Again, their product is 18. (b) x(9 – x) = 18 or, 9x – x2 = 18, which is quadratic equation. (c) x2 – 9x + 18 = 0 or, x2 – 6x – 3x + 18 = 0 or, x(x – 6) – 3(x – 6) = 0 or, (x – 3) (x – 6) = 0 So, x = 3 or 6.

196 Allied The Leading Mathematics-10 Quadratic Equation 197 Algebra Example-2 If 7 is subtracted from double of the square of a positive number, the result is 91. (a) Express the given problem in the mathematical statement. (b) Write the above statement in standard form and name the type of quadratic equation. (c) Solve this equation and write the positive number. Solution: (a) Let x be a positive number. Then, by question, 2x2 – 7 = 91 => x2 – 49 = 0 (b) The required standard form is x2 – 49 = 0, which is pure quadratic equation. (b) x2 = 49 = 72 ∴ x = ± 7 Hence, the required positive number is 7. Example-3 Among three consecutive integers, the product of the first and the third is 23 more than 10 times the second. (a) Express the given problem in the mathematical statement. (b) Write the mathematical statement in standard form and its degree. (c) Solve this equation and check the values of the variable. Solution: (a) Suppose, x, x + 1 and x + 2 represent the three consecutive integers. Then, by question x(x + 2) = 23 + 10(x + 1) ......... (i) (b) Here, x(x + 2) = 23 + 10(x + 1) or, x2 + 2x = 23 + 10x + 10 or, x2 + 2x = 10x + 33 or, x2 + 2x – 10 x – 33 = 0 or, x2 – 8x – 33 = 0, which is standard form and its degree is 2. (c) Here, x2 – 8x – 33 = 0 or, x2 – 11x + 3x – 33 = 0 or, x(x – 11) + 3(x – 11) = 0 or, (x – 11) (x + 3) = 0 ∴ x = 11, – 3 So, first number = 11, second number = 11 + 1 = 12 and third number = 11 + 2 = 13. Or, first number = –3, second number = –3 + 1 = –2 and third number = –3 + 2 = – 1. Hence, the integers we want are either 11, 12 and 13 or – 3, – 2 and – 1. Example-4 A number between 10 and 100 equals to four times of the sum of the digits and the product of the digits is 18. (a) Write the given conditions in the equation forms and combine them. (b) Solve the combined equation and find the number. Solution: (a) Let 10x + y be a two-digit number between 10 and 100, where x is in tens and y, in ones. Then by the question, For checking, putting x = 11 in eqn (i), we get x(x + 2) = 23 + 10(x + 1) or, 11(11 + 2) = 23 + 10(11 + 1) => 143 = 143 Putting x = –3 in eqn (i), we get –3(–3 + 2) = 23 + 10(–3 + 1) => 3 = 3

196 Allied The Leading Mathematics-10 Quadratic Equation 197 Algebra 1st condition, 10x + y = 4(x + y) or, 10x + y = 4x + 4y or, 10x – 4x = 4y – y or, 6x = 3y or, y = 2x ..... (i) (b) From the eqn (ii), x2 = 9 ⇒ x = 3 or –3 Since the required number lies between 10 and 100, x = – 3 is rejected. From the eqn (i), y = 2 × 3 = 6. Hence, the required number is 10x + y = 10 × 3 + 6 = 36. III Applications of Quadratic Equations on Ages The Problems of Age among two persons can be solved by using the solving quadratic equation. Now, we will remember the tricks that help us to convert the given problem into notation. Consider the present ages of two persons are x years and y years. ► The age after/later or ‘n’ years hence, their ages will be x + n and y + n. ► The age before ‘n’ years or ‘n’ years past/earlier/ago, their ages will be x – n and y – n. Example-5 The present ages of the husband and his wife are 30 and 27 years respectively. (a) What would be the ages of the husband and wife x years ago ? (b) If the product of their ages was 180, write the given condition in a notation form. (c) Find the value of x and test it in the given condition. (d) What were their ages x years ago in actuality before their marriage? Write. Solution: (a) The present ages of the husband and his wife are 30and27yearsrespectively. Suppose x years ago, their ages were (30 – x) and (27 – x) respectively. (b) If the product of their ages was 180. i.e., (30 – x) × (27 – x) = 180 (c) Here, (30 – x) × (27 – x) = 180 or, 810 – 30x – 27x + x2 = 180 or, x2 – 57x + 810 – 180 = 0 or, x2 – 57x + 630 = 0 or, x2 – 42x – 15x + 630 = 0 or, x(x – 42) – 15(x – 42) = 0 or, (x – 42) (x – 15) = 0 Either, x – 42 = 0 ∴ x = 42, which is not possible. 2nd condition, x × y = 18 or, x × 2x = 18 => 2x2 = 18 ..... (ii) which is combined equation. 630 = 2 × 3 × 3 × 5 × 7 = 42 × 15 s.t. 42 + 15 = 57

198 Allied The Leading Mathematics-10 Quadratic Equation 199 Algebra OR, x – 15 = 0 ∴ x = 15 Hence, 15 years ago, the product of their ages was 180. (d) 15 years ago, the ages of husband and wife were 30 – 15 = 15 years and 27 – 15 = 12 years before their marriage respectively. Example-6 The present age of a mother is 3 years older than 4 times his son's age. The product of their ages 11 years ago was 40. (a) What type of equation do you find from the given problem? (b) Write the given problem in a single equation in standard form. (c) Solve the equation. (d) At what age of the mother was given birth to her son ? Solution: (a) We find the quadratic equation from the given problem, (b) Let x and y be the present ages of a father and his son. Then by question, x = 4y + 3 ............. (i) or, (x – 11) (y – 11) = 40 ........... (ii) or, (4y + 3 – 11) (y – 11) = 40 [∴ From eqn (i)] or, (4y – 8) (y – 11) = 40 or, 4y2 – 44y – 8y + 88 – 40 = 0 or, 4y2 – 52y + 48 = 0 or, y2 – 13y + 12 = 0 (c) Again, for solving, y2 – 13y + 12 = 0 or, y2 – 12y – y + 12 = 0 or, y(y – 12) – 1(y – 12) = 0 or, (y – 12)(y – 1) = 0 Either, y – 12 = 0 ∴ y = 12 Or, y – 1 = 0 ∴ y = 1, which does not satisfy the given condition. So, it is rejected. From the eqn (i), x = 4 × 12 + 3 = 51 years. (d) The difference between their ages is 51 – 12 = 39 years. Hence, the mother gave birth to her son at 39 years. IV Applications of Quadratic Equations on Mensuration The solving of quadratic equations are almost used in the case of area of triangles and quadrilaterals, lateral or curved surface area, total surface area and volume of the prism, cylinder, pyramid, cone, sphere, hemi-sphere, and their combined solids. So, we should remember all the formulae studied in the unit Mensuration.

198 Allied The Leading Mathematics-10 Quadratic Equation 199 Algebra Example-7 Mohan found that the area of a square garden is 25 square meters. But, he forgot the measure of its length. (a) Which formula does he use to find the length of the garden? (b) How long the garden will he find ? (c) How much will he expense for fencing around the garden five times at the rate of Rs. 250 per meter ? Solution: (a) He uses the formula, A = l 2 , where l = length of a square and A = area of the square, to find the length of the garden. (b) Suppose the side of a square garden is x units. Then its area is x2 . Given that the area of the square garden is 25 sq. m, we have x2 = 25 = 52 ∴ x = ± 5 => x = 5 or x = – 5, rejected. Hence, he will find the length of the square garden 5 meter. (c) Length of fence at once around the garden = Perimeter of square (P) = 4l = 4 × 5 = 20 m ∴ Total length of fence at 5 times around the garden (L) = 5 × 20 = 100 m Rate of fence (R) =Rs. 250/m ∴ Expense for fence = R × L = Rs. 250 × 100 = Rs. 25000. Example-8 The black enamel painted area of the adjoining cylindrical oil tank of the length 20 meters is 517 sq. meters. (a) Which formula is used to calculate the radius of the circular base of the given tank ? (i) TSA = 2πr(r – l) (ii) TSA = 2πrh (iii) CSA = 2πr2 (iv) TSA = 2πr(r + l) (b) What is the diameter of the tank ? Find it. (c) Calculate the amount of petrol containing in the tank at Rs. 181 per litre. Solution: (a) The formula TSA = 2πr(r + l) is used to calculate the radius of the circular base of the given tank. (b) Here, length of the given cylindrical oil tank (l) = 20m Black painted area of the tank (TSA) = 517 m2 . Radius of the circular base of the tank (x) = ? Diameter of the circular base of the tank (y) = ? Now, we have TSA = 2πr(r + l) or, 517 = 2 × 22 7 x(x + 20) 20 m A = 517 m2

200 Allied The Leading Mathematics-10 Quadratic Equation 201 Algebra or, 3619 = 44x2 + 880x or, 44x2 + 880x – 3619 = 0 or, 11(4x2 + 80x – 329) = 0 or, 4x2 + 94x – 14x – 329 = 0 or, 2x(2x + 47) – 7(2x + 47) = 0 or, (2x + 47) (2x – 7) = 0 Either, 2x + 47 = 0 ∴ x = – 47 2 , rejected or, 2x – 7 = 0 ∴ x = 7 2 = 3.5 m ∴ The diameter of the tank is 2 × 3.5 = 7 m. (c) Again, volume of petrol containing in the tank = Volume of petrol in it. or, V = πr2 h or, V = 22 7 × (3.5)3 × 20 = 2695 sq. m. (d) Rate of petrol (R) = Rs. 181 per litre ∴ Volume of petrol in the tank (V) = πr2 h = 22 7 × 3.53 × 20 = 2695 m3 ∴ Amount of petrol (V) = 2695 m3 = 2695 × 1000 l = 2695000 l Hence, cost of petrol in the tank (T) = Rs. 181 × 2695000 = Rs. 48,77,95,000. V Applications of Quadratic Equations on Distance and Time Finding the distance, time or speed of a given object may lead to the formation of a quadratic equation. So, we remember some relations of Science. 1. Speed of a moving object = Distance Time 2. Speed upstream = (u - v) km/hr and speed upstream = (u + v) km/hr, where u = Speed of a moving object in still water, v = Speed of water current 3. Kinetic energy, E = 1 2 mv2 , where m = mass of an object, v = velocity of an object 4. Potential energy, PE = mgh, where g = acceleration due to gravity, h = height of the object Example-9 A body of mass 10 grams projected upwards acquires a velocity (v) at a certain point above the ground. Its kinetic energy at that point is 500 units. (a) Write the formula relating with the kinetic energy (E), mass (m) and the velocity (v) of the body. (b) Find the velocity of the body at the point observed. (c) What refers to the positive and negative values of the velocity ? Solution: (a) The formula relating with the energy (E), mass (m) and the velocity (v) of the body is E = 1 2 mv2 .