vedanta Excel In Opt. Mathematics - Book 10
Project Work
20. An employee gets starting salary of Rs. 20000 per month. If his salary is increased by of
Rs. 4000 every year, find the following:
(i) the salary of the employee in the 5th year.
(ii) the salary of the employee in the 10th year.
(iii) If his starting salary is Rs. 30000 and yearly increment is Rs. 5000, find his salary
in the 8th year.
3. (a) 2 (b) 5 (c) -2
4. (a) Yes
5. (a) 25 (b) No (c) Yes (d) Yes
7. (a) 2 + n
8. (a) 3, 30 (b) 10 6. A.S
9. (a) 26
11. (a) 128 (b) 3n + 4 (c) 3n + 3 (d) 57 - 6n
13. (a) -3, 22, -5
(b) 44, 8 (c) 14, 77 (d) 112, 40
(e) -3, 27, 0
16. (a) 2, 5, 8, 11, ...... (b) 10 10.(a) 50 (b) 33
17. (a) x = –3, –86,
20. (i) Rs. 36000 (b) 40 12.(a) Yes (b) Yes
(b) -3, 27, 0 (c) 7, 54, 117 (d) 4, -48, -12
14.(a) 6th (b) 17th 15.(a) 11 (b) 10
(b) 48, 46, 44, ........
–100, –114 (b) k = -3, t7 = -3, t8 = -44, t9 = -100
(ii) Rs. 56000 (iii) Rs. 65000
3.2 Arithmetic Mean
Let us study the following arithmetic sequences.
(a) 10, 20, 30 (b) 50, 40, 30, 20
Here, 20 is called arithmetic mean of 10 and 30 in sequence (a).
Also, 40, and 30 are called arithmetic means between 50 and 20 in sequence (b).
Definitions: If three numbers are in A.P. the middle term is called arithmetic mean and it is
denoted by A.M. in abbreviated form.
If sequence of numbers are in an A.P., the terms between the first and the last terms are
called arithmetic means (A.M.)
Arithmetic Mean (A.M) between two numbers.
Let three numbers a, m, and b be in A.P., by definition of A.P., we have,
m-a = b - m
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 51
vedanta Excel In Opt. Mathematics - Book 10
or, 2m = a+b
m= a+b
2
Here, m is the arithmetic mean between a and b.
For Example : Find arithmetic mean between 100 and 200.
Here, a = 100, b = 200
Arithmetic mean between a and 3b20=0 =a+215b0
i.e. A.M =
100 + 200 =
2
Arithmetic Means between Two Numbers.
Let a and b be the first and the last numbers of an A.P., there are n arithmetic means namely
m1, m2, ...... mn between a and b.
Then a, m1, m2 ...... mn, b, are in A.P.
The number of means = n
The total number of terms in the A.P. = n + 2.
Then, we write,
b = a + {(n+2) - 1} d
or, b - a = (n + 2 - 1) d
or, b - a = (n + 1) d
? d = b-a
n+1
Note :
In d = b-a , n is the number of means not number of terms.
n+1
Then, we write the arithmetic means as followers,
m1 = a + d, m2 = a + 2d, ........., mn = a + nd.
For example : Find 3 A.M.'s between 7 and 19.
Here, number of A.M.'s means (n) = 3
first term (a) = 7
last term (b) = 19
common difference (d) = b-a = 19 - 7 = 12 = 3
n+1 3+1 4
Let, the required means be m1, m2 and m3 .
Then, m1 = a + d = 7 + 3 = 10
m2 = a + 2d = 7 + 2 × 3 = 7 + 6 = 13
m3 = a + 3d = 7 + 3 × 3 = 7+9 = 16.
52 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
Worked Out Examples
Example 1. Find an A.M. between 7 and 21.
Solution:
Example 2. Let m be the required A.M. between 7 and 21, a = 7 and b = 21
Solution:
Then, m = a + b = 7+21 = 28 = 14.
Example 3. 2 2 2
Solution:
Find 5 A.M.’s between 9 and 33.
Example 4.
Solution: Here, the first term (a) = 9
The last term (b) = 33
The number of means (n) = 5
The common difference (d) = b-a = 33-9 = 24 = 4
n+1 5 +1 6
Let m1, m2, m3, m4 and m5 be the required means. Then,
m1 = a + d = 9 + 4 = 13
m2 = a + 2d =9 + 2 × 4 = 9 + 8 = 17
m3 = a + 3d = 9 + 3 × 4 = 9 + 12 = 21
m4 = a + 4d = 9 + 4 × 4 = 9 + 16 = 25
m5 = a + 5d = 9 + 5 × 4 = 9 + 20 = 29
Therefore, the required means are 13, 17, 21, 25 and 29.
If 3, p, q, r, 15 are in A.P., find the values of p, q, and r.
Here, the first term (a) = 3
The last term (b) = 15
The number of means (n) = 3
the common difference (d) = b-a = 15- 3 = 12 =3
n+1 3+1 4
Now, the first mean (m1) = p = a + d = 3 + 3 = 6
The second mean (m2) = q = a + 2d = 3 + 2 × 3 = 9
The third mean (m3) = r = a + 3d =3 + 3 × 3 =3 + 9 =12
Therefore, the required values of p,q,r are 6,9,12 respectively.
There are n arithmetic means between 20 and 80 such that the first mean :
the last mean = 1:3, find the value of n.
Here, the first term (a) = 20
the last term (b) = 80
the number of means = n
Then, the common difference (d) = b-a = 80-20 = n 60 1
n+1 n+1 +
The first mean (m1) = a + d =20 + n 60 1 = 20n+20+60 = 20n + 80
+ n+1 n+1
The last mean (mn) =a + nd = 20 + n60 = 20n+20+60n = 80n + 20
1 n+1 n+1 n+1
m1 3
By question m2 =
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 53
vedanta Excel In Opt. Mathematics - Book 10
20n+80 1
n+1 3
or, 80n + 20 =
n+1 (n+1) 1
20(n+4) 20(4n+1) 3
or, n+1 × =
or, n+4 = 1
4n +1 3
or, 3n + 12 = 4n + 1
or, n = 11
Therefore, the required number of mean is 11.
Example 5. In an A.P. some arithmetic means are inserted between 7 and 28. If the third
Solution: mean is 16. Find the number of means.
Here, the first term (a) = 7
The last term (b) = 28
let the number of means = n
The common difference (d) = b-a = 28 - 7 = 21
n+1 n+ 1 n+1
Now, the third mean (m3) = a + 3d
21
or, 16 = 7 + 3. n+1
or, 16 - 7 = 63
n+1
7
or, 1 = n + 1
or, n + 1 = 7
? n=6
Hence, there are 6 means.
Exercise 3.2
Very Short Questions
1. (a) Define arithmetic means.
(b) Write arithmetic mean between p and q.
(c) Write the formula of finding common difference of A.P. when the first term = a
and last term = b, and number of means = n
(d) If common difference = d, the first term = a, the last term = b, of an A. P. find, the
three means in term of a and d between a and b.
2. Find arithmetic mean between the following two terms.
(a) 40 and 60 (b) -60 and -20 (c) 7 and 11
3 3
3. (a) If A.M. between 2k + 3 and 5k + 8 is 6, find the value of k.
(b) If A.M. between 3x + 7 and 5x + 8 is 23 , find the value of x.
2
54 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
Short Questions
4. Find the arithmetic means as indicated
(a) 5 A.M.’s between 2 and 20.
(b) 3 A.M.’s between 6 and 11.
(c) 4 A.M.’s between 140 and - 60.
(d) 3 A.M.’s between -11 and 4.
5. (a) Find the value of x if 10, x and 30 are in A.P.
(b) If 6,x,y, and 18 are in A.P., find the value of x and y.
(c) If 96, p, q,r, 68 are in A.P., find the values of p,q and r.
(d) If x, x2 + 1 and x + 6 are in A.P., find the value of x.
Long Questions
6. (a) 6 arithmetic means between 4 and p are inserted and the fifth mean is 14, find the
value of p. Also, find the means.
(b) There are 3 A.M.’s between a and b. If the first mean and third mean are 10, 20
respectively, find the values of a and b.
7. (a) The 5th mean between two numbers 7 and 71 is 27. Find the number of A.M’s.
(b) There are n arithmetic mean between 5 and 35. If the second mean to the last
mean is 1:4, find n.
(c) There are n arithmetic mean between 1 and 70. If the first mean : the last mean =
4 : 67, find n.
(d) Find the number of arithmetic means between 2 and 37 where the second mean :
last mean = 3 : 8.
1. (b) m = p + q (c) d = b-a (d) a + d, a + 2d, a + 3d
2 n+1
1
2. (a) 50 (b) -40 (c) 3 3.(a) 7 (b) 1
4. (a) 5, 8, 11, 14, 17 (b) 29 , 17 , 39 (c) 100, 60, 20, -20 (d) - 29 , - 7 , - 1
5. (a) 20 4 2 4 (c) 89, 82, 75 4 2 4
(b) 10, 14 (d) -1, 2
6. (a) P = 18, Means = 6, 8, 10, 12, 14, 16 (b) a = 5, b = 25
7. (a) 15 (b) 17 (c) 22 (d) 6
3.3 Sum of Arithmetic Series
A TV manufacturing company manufactured the number television as follows.
Year 2011 2012 2013 2014 2015 2016
TV 10000 20000 30000 40000 50000 60000
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 55
vedanta Excel In Opt. Mathematics - Book 10
Then answer the following questions.
(a) Is the sequence of production of TV in A.P. ?
(b) Find the sum of TV’s for the six years as shown in the table.
(c) If STnV=inn2s[ixa+yle]a, rns = 6, a = 10000, l = 60000, find S6. Does s6 gives the total production
of ?
(d) If l = a + (n - 1) d, can we write Sn = n [2a + (n-1)d]?
2
To find the formula of sum of n terms of an A.P.
Let ‘a’ and d denote the first term and common difference of an A.P. with n terms l the last
term. Then a, a + d, a + 2d, a + 3d,...... l - 2d, l - d, l are in A.P.
Let Sn denote the sum of n terms of the A.P.
Now, Sn = a + (a + d) + (a + 2d) + ...... + (l - 2d) + (l - d) + l
Writing the terms in reverse order.
Sn = l + (l -d) + (l - 2d) + ..... + (a + 2d) + (a + d) + a
Adding (i) and (ii) sums, we get
2Sn = (a + l) + (a + l) + (a + l) + ....... + (a + l)
= Sum of n times (a + l)
= n (a + l)
? Sn = n (a + l)
2
But, l = a + (n - 1) d, we get
Sn = n [2a + (n - 1)d]
2
Note :
(i) If a, n and l of AP. are known, we use Sn = n [a + l]
2
n
(ii) If a, n, and d of an A.P. are known, we use Sn = 2 [2a + (n-1)d]
Worked Out Examples
Example 1. Find the sum of the first 10 terms of the series 1 + 4 + 7 + 10 + .......
Solution:
This is an A.P.
Common difference (d) = 4 - 1 = 3
The first term (a) = 1
The number of terms (n) = 10
S10 = ? n
2
By using formula, Sn = [2a + (n-1)d]
56 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
or, S10 = 10 [2.1 + (10-1)3] = 5 [2 + 9.3]
2
= 5.[2 + 27] = 5.29 = 145
Hence, the sum is 145.
Example 2. 10
Solution:
Evaluate : ∑ (n + 4)
Exercise 3.
Solution: n=5
Example 4. 10
Solution:
Here, ∑ (n + 4) = (5+4)+ (6+ 4) + (7 + 4) + (8 + 4) + (9 + 4) + (10 + 4)
n=5 = 9 + 10 + 11 + 12 + 13 + 14 = 69
Alternative method
Here, 10 (n + 4) = (5 + 4) + (6 + 4) + (7 + 4) +....... + (10 + 4)
=9 + 10 +11 + ...... + 19
∑
n=5
This is a A.P. with a = 9, d = 10 -9 = 1, n = 6
S6 = ?
By using formula,
n
Sn = 2 [2a + (n - 1)d]
or, S6 = 6 [2.9 + (6 - 1).1] = 3[18 + 5] = 3.23 = 69
2
Hence, the sum is 69.
From the series 5 + 10 + 25 +...... + 100, find the number of terms in the
series and Sn.
Here, 5 + 10 + 15 + ......... + 100
This is an AP with a = 5, d = 10 - 5 = 5
Let tn = 100, we have,
or, a + (n - 1) d = 100
or, 5 + (n-1) 5 = 100
or, 5 + 5n - 5 = 100 or, 5n = 100
? n = 20
n
Now, Sn = 2 [a + l]
or, S20 = 20 [5 + 100] = 10.105 =1050
2
Therefore, n = 20, and S20 = 1050.
Find the sum of the first 8 terms of an A.P. whose first term is 15 and the
third term is 1.
In a given A.P.
The first term (a) = 15
The third term (t3) = 1
By using formula, we get
t3 = a + (n-1).d
or, t3 = 15 + (3-1).d
or, 1= 15 + 2d
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 57
vedanta Excel In Opt. Mathematics - Book 10
or, 2d = -14
? d = -7
Now, using formula,
Sn = n [2a + (n-1)d]
or, S8 2
8
= 2 [2.15 + (8-1).(-7)]
= 4[30 + (7).(-7)] = 4[30 - 49]
= 4.(-19) = -76
Hence, the sum is -76.
Example 5. Find the common difference of an A.P. whose first term and sum of the first
15 terms are 100 and 450 respectively.
Solution: In given A.P., a =100, S15 = 450
By using formula,
Sn = n [2a + (n - 1)d]
2
15 or, 450 = 125[200 + 14d]
or, 450 = 2 [2.100 +(15-1)d] or, - 140 = 14d
or, 60 = 200 + 14d
? d = -10
Hence, the common difference of the AP is -10.
Example 6. How many terms of an A.P. 24, 20, 16,..... must be taken so that the sum may
be 72? Explain the double answer.
Solution: Here, in given A.P.
a = 24, d = 20 - 24 = - 4, Sn = 72. n = ?
By using formula,
Sn = n [2a + (n -1)d]
or, 2 n [2 × 24
2 n
72 = + (n - 1).(-4)] or, 72 = 2 [48 - 4n + 4]
or,
or, 144 = n[52 - 4n] 144 = 4n[13 - n]
or, 36 - 13n - n2 = 0 or, n2 - 13n + 36 = 0
or, n2 - 9n - 4n + 36 = 0 or, n(n-9) - 4 (n - 9) = 0
or, (n-9) (n-4) = 0
? n = 4 or 9.
In the given A.P., common difference is negative. There are some terms in the
series whose sum is zero. The first four terms gives the sum 72. The next five
terms give the sum zero. Check,
S4 = 4 [2.24 + (4 - 1). (-4)] = 2[48 - 12] = 72
2
9 9
and S9 = 2 [2.24 + (9-1)(-4)] = 2 [48 - 32]
= 9 . 16 = 72
2
Hence, the double answer is possible and correct.
58 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
Example 7. The sum of the three numbers in A.P. is 15 and the sum of their squares is 83.
Solution: Find the numbers.
Example 8. Let the three numbers in A.P. be a-d, a and a + d.
Solution:
Now by question, a - d + a + a + d = 15
or, 3a = 15 ? a=5
Again, by question,
(a-d)2 + a2 + (a +d)2 = 83
or, (5-d)2 + 52 + (5+d)2 = 83
or, 25 - 10d + d2 + 25 + 25 + 10d + d2 = 83
or, 2d2 = 83 - 75
or, 2d2 = 8
or, d2 = 4 =22
? d = ±2
When d = 2, the required three numbers are given by a - d = 5 - 2 = 3, a = 5,
and a + d = 5 + 2 =7
when d= -2, the requires three numbers are given by
a - d = 5 -(-2) = 7, a = 5, a + d = 5 - 2 = 3.
Hence, the required three numbers are 3,5,7 or 7, 5, 3.
Find the sum of the numbers between 1 and 100 which are exactly divisible
by 3.
The sequence of numbers is 3, 6, 9, 12,.... 99 which are exactly divisive by 3.
Then the sequence is arithmetic.
a = 3, d = 3, tn = 99 (if possible)
tn= 99
or, a + (n - 1) d =99
or, 3 + (x - 1) d = 99 or, 3 + (n - 1) 3 = 99
Dividing both by 3, we get,
or, 1 + (n - 1) = 33
? n = 33
Now, S33 = 33 [2.3 + (33 - 1). 3]
2
[ ]? Sn
= 33 [6 + 96] = n n + (n - 1)d
2 2
33
= 2 [102] = 1683
Alternative Method,
S33 33 99], [ ]Snn
= 2 [3 + = 2 (a + l) = 1683
= 33 × 102 = 1683
2
Hence, the sum is 1683.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 59
vedanta Excel In Opt. Mathematics - Book 10
Note :
(a) Denote three numbers in A.P a-d, a, a+d
(b) Denote four numbers in A.P. a - 3d, a - d, a + d, a + 3d.
(c) While calculating for the value of number of terms n, accept the positive integral values
of n, reject the fractional and negative values of n. If two values of n are positive integer
check the sum, and accept the correct value of n whose n terms sum is true.
Exercise 3.3
Very Short Questions
1. (a) Write the formula of sum of n terms of an A.P. if a, d, and n of an A.P. are given.
(b) Write the formula of sum of n terms of an A.P. whose n, a and l are given.
2. Find the sum of sum of n terms of an A.P. under the following conditions.
(a) - 10 -8 ........ (4 terms)
(b) -5 -2 + 1 + 4 + ....... (8 terms)
(c) 2 + 6 + 8 + 10 ........ (10 terms)
(d) 10 + 20 + 30 + ......... (8 terms)
(e) 4 + 9 + 14 + .......... (8 terms)
Short Questions
3. Evaluate the following :
10 5 8
(a) ∑ (2n + 1) (b) ∑ (3n + 1) (c) ∑ (2n - 1)
n=1 n=4 n=2
4. Find the sum of the first 10 terms under the following conditions.
(a) a = 5, d = 2 (b) a = 20, d = -5,
(c) a = 25, d = 5 (d) a = 100, d = 20
5. (a) An arithmetic series has 10 terms. If its last term is 50 and the sum of its terms is
275, find the first term.
(b) An arithmetic series has 9 terms and the sum is 135. Find the last term if the first
term is 3.
6. (a) Find the number of terms in an A.P. which has its first term 16; common difference
4; and the sum, 120.
(b) Find the number of terms in an A.P. with the first term 2, and common difference
2 sum 420.
7. (a) Find the common difference of an A.P. whose first term is 5 and the number of
terms is 30 and sum 1455.
(b) Find the common difference of an A.P. whose first term is 9, the number of terms
is 7, and the sum is 0.
60 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
Long Questions
8. (a) How many terms of the series 2 + 4 + 6 + ..... amount to 420 ?
(b) Find the number of terms of the series 2 + 4 + 6 + ...... in order that the sum may
be 240. Explain the double answer.
9. (a) If the 9th and 29th terms of an A.P. are respectively 40 and 60, find the common
difference, the first term, sum of the first 30 terms.
(b) The fifth and twelve terms of an A.P. are respectively 17 and 45. Find the sum of
the first 15 terms of the progression?
(c) The sum of the first ten terms of an A.P. is 520. If the 7th terms is double of the third
term, calculate the first term, common difference, and sum of 20 terms.
10. (a) The sum of the first 6 terms of an A.P. is 42 and the ratio of the 10th and the 30th
term is 1:3, find the first term and the thirteenth term of the A.P.
(b) The sum of the first ten terms of an A.P. is 50 and its 5th term is treble of the second
term. Find the first term, common difference and sum of the first 20 terms.
11. (a) Find three numbers in A.P. whose sum is 9 and the product is - 165.
(b) Find the three numbers in an A.P. whose sum is 15 and their product is 80. Find them.
(c) Find the three numbers in an A.P. whose sum is 12 and the sum of their squares is 50.
12. A firm produced 1000 sets of radio in the first year. The total number of radio sets
produced at the end of 10 year is 14500.
(a) Estimate by how many units of production increased each year.
(b) On the basis of the estimate of the annual increment in production, forecast level
of output for the 15th year.
13. A person pays a loan of Rs. 3900 in monthly installments, each installment being less
than the former by Rs. 20. The amount of the first installment is Rs. 400. In how many
installments is the entire amount paid off? Give reason.
1. (a) Sn = n [2a + (n - 1)d] (b) Sn = n [a + l] 2. (a) -28
2 2 (e) 172
(b) 44 (c) 110 (d) 360
3. (a) 120 (b) 29 (c) 63
4. (a) 140 (b) -25 (c) 475 (d) 1900
5. (a) 5 (b) 27 6.(a) 5 (b) 20
7. (a) 3 (b) - 3 8.(a) 20 (b) 15
9. (a) a = 32, d = 1, S30 = 1395 (b) 435 (c) a = 16, d = 8, S20 = 1840
10. (a) a = 2, t13 = 26 (b) a = 12, d = 1 , S20 = 200
11. (a) -5, 3, 11 or 11, 3, -5 (b) 2, 5, 8 or 8, 5, 2
(c) 3, 4, 5 or 5, 4, 3 12. (a) 100 (b) 2400 13. 15
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 61
vedanta Excel In Opt. Mathematics - Book 10
3.4 Geometric Sequence and series
Let us study the following sequence of numbers.
(i) 5, 10, 20, 40, 80, .......
(ii) 1, 7, 49, 343, 2401,.......
In above sequences, each number is obtained by multiplying the preceding term by a
constant number.
The ratio of any term to its preceding term is constant. In sequence (i),
10 = 2, 20 = 2, 40 = 2, 80 =2,.......
5 10 20 40
Here, constant ‘2’ is common ratio. It is denoted by ‘r’.
In sequence (ii),
7 = 7, 49 = 7, 343 = 7, 2401 = 7,......
1 7 49 243
Common ratio in sequence (ii) is 7.
The corresponding series of above sequences are given by,
5 + 10 + 20 + 40 + 80 + ......
1 + 7 + 49 + 243 + 2401 + ....
Definition: A sequence of numbers is said to be geometrical sequence or geometric
progression (G.P) if the ratio of a term with its preceding term is constant throughout the
whole sequence. The constant is called common ratio and it is denoted by r.
or, r= tk + 1 ,k N. etc.
r= tt12tk, r = t3
t2
In a geometric progression each term after the first term is obtained by multiplying the
preceding term by a constant number. This constant number is called common ratio.
General Term of a Geometric Sequence :
Let a be the first term, r the common ratio of a geometric sequence. In a geometric sequence
each term after the first term is obtained by multiplying by common ratio r. Then a, ar2, ar3,
ar4, ...... are terms of a G.S.
t1 = a = ar° = ar1-1
t2 = ar = ar2-1
t3 = ar2 = ar3-1
? tn = arn-1
General term means the term from which we can get any required term of the sequence.
62 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
Worked Out Examples
Example 1. Show that each of the following sequence of numbers are in G.P.
(a) 1, 4, 16, 64, ..... (b) 3, 6, 12, 24, 48,.......
Solution: (a) Here, t2 = 4 = 4
t1 1
t3
t2 = 16 = 4
4
t4 64
t2 = 16 = 4 ? Common ratio = 4
Therefore, the common ratio is constant throughout the whole sequence.
Hence, the given sequence is geometric.
(b) Here, t2 = 6 = 2
tt31 = 3
t2
12 = 2
6
t4 24
tt35 = 12 = 2
t4 =
48 = 2 ? Common ratio = 2
24
Therefore, the common ratio is constant throughout the whole sequence
and the given sequence is geometric.
Example 2. Find the three terms of a G.P. whose common ratio is 5 and the first term is 2.
Solution: Here, the first term (t1) = 2
Example 3. The common ratio (r) = 5
by using formula,
tn = arn-1
t2 = 2.52-1 = 2 × 5 = 10
t3 = 2.53-1 =2.52 = 2.25 = 50
Find the common ratio and the 10th term of the sequence 2+6+18+54 ......
Solution: Here, the first term (t1) =(a) = 2
The second term (t2) = 6 t2 6
t1 2
The common ratio (r) = = =3
Given sequence is a geometric;
Now, tn = arn-1
t10 = 2.310-1 = 2.39 = 39366
Example 4. Find the number of terms in the sequence. 3 - 6 + 12 - 24 + ...... - 1536
Solution: Here, the first term (a) = 3
The common ratio (r) = t2 = -6 = -2
t1 3
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 63
vedanta Excel In Opt. Mathematics - Book 10
Now, given series is geometric
tn = arn-1
or, -1536 = 3 (-2)n-1
or, -512 = (-2)n-1
or, (-2)9 = (-2)n-1
or, 9 = n - 1
? n = 10
Therefore, G. S. has 10 terms.
Example 5. In a G.P. if the common ratio is 2 and the 10th term is 1536, find the first term.
Solution:
Here, common ratio (r) = 2
Example 6.
Solution: t10 = 1536
Now, tn = arn-1
or, 1536 = a.210-1
or, 1536 = 512a
? a = 3
Therefore, the first term is 3.
From the given geometric series 4 + 2 + 1 + 1 + ...... + 1
2 4 512
(i) Find the number of terms.
(ii) IWs h61i4chatteerrmmiosf2t15h6e ?
(iii) series?
(i) Here, a = 4
common ratio (r) = t2 = 2 = 1
t1 4 2
1 1
Let tn = 512 or, arn-1 = 512
( )or,4. 1 n-1 = 1
2 512
n-1
( )or, 1 = 1 = 1
2 512 u 4 2048
( ) ( )or,1 1 11
2 n-1 = 2
or, n – 1 = 11
? n = 12
Hence, the number of terms is 12.
(ii) Again, let tn = 1
256
( )or, 18
arn – 1 = 2
( ) ( )or, 1 18
4. 2 n–1= 2
64 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
( ) ( )or,1 1 10
2 n–1= 2
? n – 1 = 10
or, n = 11
Hence, 1 is the eleventh term of the G.P.
Here, a =2546
(iii)
common ratio (r) = 1
Let, tn = 2
1 if possible, then find n.
64
we know that
or, 1( )tn = arn-1=4 1 n-1
64 2
( )or, 1 = 1 n-1
256 2
( ) ( )or, 8 1 n-1
1 = 2
2
? 8 = n - 1, or, n = 9
Yes, it is the 9th term.
Example 7. If the 4th and the 8th terms of a G.P. are 24 and 384 respectively, find the 12th term.
Solution: Let ‘a’ and r be the first term and common ratio.
Example 8.
t4 = 24
t8 = 384
By formula, tn = arn – 1, we get
or, t4 = ar3
or, 24 = ar3 .......... (i)
Again t8 = ar7
or, 384 = ar7 .......... (ii)
Dividing (ii) by (i), we get
384 = ar7
24 ar3
or, 16 = r4 or, r4 = 24
? r =2
Putting the value of r in (i), we get
24 = a.23
? a = 3
Now, t12 = ar11 = 3.211 = 6144
Find the geometric series whose, 5th term and 8th term are 80 and 640
respectively.
Solution: Let a and r denote the first term and the common ratio of the given G.P.
Then, t5 = ar 5 - 1
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 65
vedanta Excel In Opt. Mathematics - Book 10
or, 80 = ar4 .............. (i)
and t8 = ar 8 - 1
or, 640 = ar7 .............. (ii)
Dividing (ii) by (i), we get,
68400= ar7
ar4
or, 8 = r3 or r3 = 23
? r =2
Putting the value of r in equation (i)
80 = a.24
or, 80 = 16a
? a = 5
Required geometric series is given by,
a + ar + ar2 + ...
= 5 + 5.2 + 5.22 + ...
= 5 + 10 + 20 + ...
Example 9. In a G.P. 7th term is 81 times the 3rd term and the 5th term is 243. Find the series.
Solution: Let a and r denote the first term and common ratio respectively.
Now, t5 = ar4 [ tn = arn – 1]
or, 243 = ar4 ............ (i)
By question,
t7 = 81 u t3
or, ar6 = 81.ar2
or, r4 = 81 or, r4 = 34
? r=3
From equation (i),
243 = a.33
or, 243 = a.81 or, a = 3
The required series is given by,
a + ar + ar2
= 3 + 3.3 + 3.32 + ...
= 3 + 9 + 27 + ...
Example 10. Find the three numbers in G.P. whose sum is 14 and product is 64.
Solution:
Let three required terms in G.P. be a, a and ar.
. ar = 64 r
By question, a. a
r
or, a3 = 64 = 43
? a=4
Again, by question,
a + a + ar = 14
r
66 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
or, 4 + 4 + 4r = 14
r
or, 4 + 4r + 4r2 = 14r
or, 4r2 – 10r + 4 = 0
or, 2r2 – 5r + 2 = 0
or, 2r2 – 4r – r + 2 = 0
or, 2r (r – 2) – 1 (r – 2) = 0
or, (r – 2) (2r – 1) = 0
? r = 2 or 1
2
When r = 2 , the required numbers are given by
a = 4 = 2, a = 4 and ar = 4.2 = 8
r 2
When r = 21, the required numbers are given by
a 4 4.21
r = 1\2 = 8, a = 4 and ar = = 2
Therefore, the required numbers are 2, 4, 8, or 8, 4, 2.
Exercise 3.4
Very Short Questions
1. (a) Define a geometric series.
(b) Define common ratio in G.P.
(c) Write general term formula in a G.P.
(d) In a G.P., tn = arn – 1, write the meanings of a, r and n.
2. Check the following sequences are in G.P. or not.
(a) 2, 4, 8, 16, ... (b) 3, 9, 27, 81, ...
(c) –2, –4, – 12, 16, ... (d) 1, 31, 19, 217, 2143, ...
3. Find the common ratio in the following G.P.
(a) 2, 4, 8, 16, ... (b) 1 + 1 + 1 + 2 + 4 + ...
4 2
1 1 1 1
(c) 1 – 3 + 32 + ... (d) 2+ 2 + 22
4. Find the next three terms in each of the following cases of G.P.
(a) 2, 4, 8, 16,... (b) 3, 9, 27, 81,...
(c) 1 + 1 + 1 + 1 + ... (d) 1 – 1 + 1 – 1 + ...
3 9 27 4 16 64
Short Questions
5. Find the 5th and the 10th terms of G.P. in the following cases.
(a) a = 4, r = 1 , (b) t1 = –4, r = 2
2
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 67
vedanta Excel In Opt. Mathematics - Book 10
(c) First term = 5, common ratio = 2 (d) First term = 7, second term = 14
6. (a) In a G.P. 7, 14, 28, 56, ... 1792, find the number of terms.
(b) Which term of the series 1 + 1 + 1 + ... is 243?
27 9 3
64
(c) If first term is 27, and the second term is 18. Which term will be 27 ?
(d) Find the number of terms in G.P. 5, –15, 45, ..., – 10935
7. (a) If 2, x, 8, y are in G.P. find the values of x and y.
(b) If 3, k, 27, p are in G.P. find the value of k and p.
Long Questions
8. (a) In a G,P., the first term and second terms are respectively 4 and 16, find the 10th term.
(b) In a G.P., the first term is 1 and third term is 217, find the 6th term.
3
9. (a) If 5th, 8th terms of a G.P. are 81 and 2187 respectively, find the first term, common
ratio of G.P.
(b) If 6th and 13th terms of a G.P. are 64 and 8192, find the first term and common ratio.
10. (a) If the 3rd and the 6th terms of a G.P. are 36 and 243 respectively. Then, find:
2
(i) the first term (ii) the common ratio (iii) the 10th term
(b) The first and the second terms of a G.P. are 32 and 8 respectively. Find:
(i) the first term (ii) common ratio (iii) 10th term
11. (a) If the 6th and the 9th terms of a G.P. are 1 and 8 respectively. Which term of the
sequence is 64?
(b) Which term of the sequence is 1 if its 3rd and 5th terms are 27 and 3 respectively.
9
1 811 .
12. (a) Find the geometric series whose the 3rd and the 6th terms are respectively 3 and
(b) Find the geometric series whose 4th and 9th terms are respectively 1 and 1128.
4
13. (a) In a G.P. the 5th and 8th terms of are respectively 256 and 32 which term is 2?
(b) In a G.P., the 2nd and the 5th terms are respectively 32 and 4. Which term is 1?
14. (a) The third term of a G.P. is 27 times the 6th term and the 4th term is 9. Find the
series.
(b) In a G.P., the 7th term is 16 times the 3rd term and the 5th term is 1 . Find the 3rd
term? 16
15. (a) The sum of three numbers in G.P. is 38 and their product is 1728. Find the numbers.
(b) Sum of three consecutive terms in G.P. is 28 and their product is 512. Find the
numbers.
(c) The product of three numbers in G.P. is 729 and sum of their squares is 819. Find
the numbers.
68 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
1. (c) tn = arn - 1 2.(a) Yes (b) Yes (c) Not
(d) Not 3.(a) 2 (b) 2 (c) - 1
3
1 1 1 1
(d) 2 4.(a) 32, 64, 128 (b) 243, 729, 2187 (c) 81 , 243 , 729
(d) 1 , - 1 , 1 5.(a) 1 , 1 (b) -64, -2048
256 1024 4096 4 128
(c) 80, 2560 (d) 112, 3584 6.(a) 9 (b) 9
(c) 7 (d) 8 7. (a) 4, 16 (b) 9, 81
8. (a) 1048576 (b) 729 9.(a) 1, 3 (b) 2, 2
10. (a) 16, 3 , 19683 (b) 32, 1 , 1 11.(a) 12th (b) 8th
2 32 4 8192
1 1
12. (a) 3 + 1 + 3 ..... (b) 2 + 1 + 2 ...... 13. (a) 12th (b) 7th
14. (a) 243 + 81 + 27 + ......... (b) 1
64
15. (a) 8, 12, 18 or 18, 12, 8 (b) 4, 8, 16 or 16, 8, 4 (c) 27, 9, 3
3.5 Geometric Mean (G.M.)
Let us consider the following geometric sequence:
(a) 3, 9, 27 (b) 20, 40, 80, 160
In above sequences (a) 9 is in between 3 and 27. It is a geometric mean (G.M.) between 3
and 27.
In sequence (b), 40 and 80 are geometric means (G.M.) between 20 and 160.
Definition : It three numbers are in G.P., then the middle term is called the geometric mean
(G.M.) between the other two.
Example: In G.P. 2, 4, 8, the geometric mean is 4, between 2 and 8.
If any number of terms are in G.P., the number of terms between the first term and last term
are called the geometric means (G.M.’s) between them.
Example: In G.P. 2, 10, 50, 250, 1250, here, 10, 50, 250 are three geometric mean (G.M.’s)
between 2 and 1250.
Geometric Mean between Two Numbers
Let G be the geometric mean between a and b. Then a, G, b are in G.P.
Then, by definition of G.P,
G = b
a G
? G.M. = ab or, G2 = ab
or, G = ab
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 69
vedanta Excel In Opt. Mathematics - Book 10
Example: Find the geometric mean 3 and 27.
Solution:
Let G.M. be geometric mean between 3 and 27.
Here, a = 3, b = 27
G. M. = ab = 3 . 27 = 81 = 9
Geometric Means between Two numbers
Let m1, m2, m3, ... mn be n G.M.’s between a and b. Then a, m1, m2, m3, ... mn, b are in G. P.
Then, the total number terms = n + 2.
The last term b = arn + 2 – 1
or, b = rn + 1
a
( )? r= b 1
a n+1
Then, the means m1, m2, m3, ..., mn are given by,
m1 = ar, m2 = ar2,
m3 = ar3. ... mn = arn
Example: Find the three G.M.’s between 3 and 243.
Solution: Let m1, m2, and m3 be three required means then number of means (n) = 3
( ) ( )The common ratio (r)
= b 1 243 1
a 3 3+1
n+1 =
11
= (81)4 = (34)4 = 3
Now, m1 = ar = 3.3 = 9
m2 = ar2 = 3.32 = 27
m3 = ar3 = 3.33 = 81
Relation between Arithmetic Mean (A. M.) and Geometric Mean (G. M.) between two
positive numbers.
Let a and b be two unequal positive numbers and A. M. and G. M. be arithmetic mean and
geometric mean between them.
Then, A. M. = a + b
2
G. M. = ab
Now, A. M. – G. M. = a + b – ab = a + b – 2 ab
2 2
= 1 [( a )2 – 2. a. b + ( b)2]
2
1
= 2 ( a– b)2
Since the square of difference of two unequal real number is always positive,
( a – b)2 > 0
? A. M. – G. M. > 0
70 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
i.e. A. M. > G. M.
Therefore, the arithmetic mean is greater than geometric mean of two real numbers.
Example : Find A. M. and G. M. of 4 and 16, also verify A. M. > G. M.
Solution:
A. M. = 4 + 16 = 20 = 10
2 2
G. M. = 4.16 = 64 = 8
Here, 10 > 8
i.e. A. M. > G. M.
Worked Out Examples
Example 1. Find the geometric mean between 20 and 80.
Solution:
Here, a = 20, b = 80
Example 2. G. M. = ab = 20 u 80 = 1600 = 40
Solution:
If 8, x, y and 64 are in G. P., find the values of x and y.
Here, 8, x, y and 64 are in A. P.
The first term (a) = 8
The last term (b) = 64
The number of means (n) = 2
( ) ( )= 1
The common ratio (r) b 1 64 2+1
a 8
n+1 =
11
= (8)3 = (23)3= 2
Now, the first mean (m1) = x = ar = 8.2 = 16
The second mean (m2) = y = ar2 = 8.22 = 32
Therefore, the required values of x and y are 16, 32.
Example 3. Insert three geometric means between 1 and 4.
Solution: Here, 4
1
the first term (a) = 44
The last term (b) =
The number of means (n) = 3
The common ratio (r) = ( b )n 1 1 = ( 4 ) 3 1 1
a + 1/4 +
= (16)14 = (24)41 = 2
Let required G. M.’s be m1, m2, and m3 respectively.
1 1
m1 = ar = 4 .2 = 2
m2 = ar2 = 1 .22 = 1
4
1 1
m3 = ar3 = 4 . 23 = 4 . 8 = 2
Therefore, the required geometric means are 1 , 1, 2.
2
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 71
vedanta Excel In Opt. Mathematics - Book 10
Example 4. Some geometric means are inserted between 5 and 80. Find the number of
Solution: means between them if the second mean is 20.
Here, a = 5, b = 80
the second mean (m2) = 20
( ) ( )= 1
the common ratio (r) b 1 80 n+1
a 5
n+1 =
14
= (16)n + 1 = 2n + 1
But, m2 = ar2
( )4 2
20 = 5. 2 n + 1
8
or, 4 = 2n + 1
8
or, 22 =2n + 1
? 2 = n +8 1
or, 2n + 2 = 8
Example 5.
Solution: or, 2n = 6
? n = 3.
Therefore, the required number of means is 3.
There are n G. M.’s between 1 and 256. If the first mean :
4
last mean = 1:256; find the value of n.
Here, the first term (a) = 1
4
the last term (b) = 256
( ) ( )the common ratio (r) = 1
b 1 256 n+1
a 1/4
n+1 =
=(210)n 1 1 = 10
+
2n + 1
the first mean (m1) = ar
the last mean (mn) = arn
By question,
m1 1
marn = 256
arn
or, = 1
256
1 1
or, rn – 1 = 256
or, rn – 1 = 256
10 (n – 1)
or, 2 n + 1 = 28
? 10n – 10 = 8
n+ 1
72 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
or, 10n – 10 = 8n + 8
or, 2n = 18
Example 6. ? n=9
Solution: Hence, the required number of means is 9.
Find the two numbers whose arithmetic mean is 25 and geometric mean is 20.
Let two required numbers be a and b.
Then, A. M. = a + b
2
a + b
25 = 2
or, 50 = a + b
or, a + b = 50 ............... (i)
Also, G. M. = ab
or, 20 = ab
or, ab = 400 .................. (ii)
Now, a – b = (a + b)2 – 4ab
= (50)2 – 4 . 400
= 2500 – 1600 = 900
= 302 = 30
a – b = 30 .............. (iii)
Adding (i) and (iii), we get
2a = 80
or, a = 40
Putting the value of a in equation (ii), we get
? b= n44u00m0 b=er1s0are 40 and 10.
Two
Exercise 3.5
Very Short Questions
1. (a) Define geometric mean.
(b) Write the geometric mean between p and q.
(c) Write the formula for finding common ratio of a.G.P. whose first term = a, last
mean = b. number of mean = n.
(d) If common ratio = r, first term = a, last term b. Write three G.M.’s between a and
b in terms of a and r.
2. Find the geometric mean between two numbers:
(a) 10 and 1000 (b) - 9 and -16
(c) 8 and 1 (d) 1 and 1
64 3 27
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 73
vedanta Excel In Opt. Mathematics - Book 10
Short Questions
3. Find the values of x and y if the given sequences are in G. P.:
(a) 125, x, y, 1 (b) 2 , x and 3 (c) 5, x, y. 40
3 2
4. Find the values of x, y, and z in the following G.P.
(a) 1 , x, y, z, 1. (b) 3, x, y, z, 48
16
81 1 1
(c) 2, x, y, z 8 (d) 27 , x, y, z, 3 .
5. (a) Find the two G.M.’s between 5 and 625.
(b) Find the two G.M’s between 1 and 4.
2
Long Questions
6. (a) Find three G.M.’s between 1 and 9.
9
1
(b) Find three G.M.’s between 4 and 64.
(c) Insert four G.M.’s between 3 and 729.
(d) Insert four G.M.’s between 1 and 32.
7. (a) The third geometric mean between 27 and 1 is 1. Find the number of mean.
27
(b) There are three G. M. between p and q. If the first mean and the third mean are
respectively 1 and 25, find the values of p and q.
8. (a) There are three geometric means between a and b. If the first mean and third mean
1 1
are 4 and 64 , find the values of a and b.
(b) There are three means between 1 and 312. If the second mean is 1 , Find the third
mean. 2 4
9. (a) Some G.M.’s are inserted between 4 and 128. Find number of means where the
ratio of the first mean and last mean is 1:8.
(b) There are n G.M’s between 1 and 25. If the ratio of the first mean to the third
25
mean is 1:25. Find the value of n.
10. (a) Find the numbers whose arithmetic and geometric means are respectively 10 and
8.
(b) Find the two numbers whose A.M. and G.M. are respectively 25 and 20.
(c) Find the two numbers whose A.M. and G.M. are respectively. 13 and 12.
(d) Find the two numbers whose A.M. is 34 and G.M. is 16.
11. (a) The ratio of two numbers is 1:16; and their geometric mean is 1 , find the numbers.
4
(b) The ratio of two numbers is 1:81, and their geometric mean is 1, find the numbers.
74 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
1. (b) pq (c) r= b 1 (d) m1 = ar, m2 = ar2, m3 = ar3
n+1
a 1 1
22 9
2. (a) 100 (b) 12 (c) (d)
3. (a) 25, 5 (b) 1 (c) 10, 20 (d)
4. (a) 1 , 1 , 1 (b) 6, 12, 24 (c) 3, 9 , 27 (d) 3 , 1 , 3
8 4 2 2 4 27 9 9
1
5. (a) 25, 125 (b) 1, 2 6.(a) 3 , 1, 3 (b) 1, 4, 16
(c) 9, 27, 81, 243 (d) 2, 4, 8, 16 7.(a) 5 (b) p = 1 , q = 125
5
1 1
8. (a) a = 1, b = 256 (b) 16 9.(a) 4 (b) 3
10. (a) 4, 16 (b) 10, 40 (c) 8, 18 (d) 4, 64
11. (a) 1 , 2 (b) 1 , 9
8 9
3.6 Sum of Geometric Series
Let us consider a geometric series whose first term is a common ratio r, number of terms n,
last term tn. Let Sn denote the sum of n terms of the series. Then, we have,
Sn = t1 + t2 + t3 + ...... + tn
or, Sn = a + ar + ar2 + ...... + arn-2 + arn-1 ........ (i)
Multiplying both sides by r, we get,
rSn = ar + ar2 + ar3 + ....... + arn-1 + arn ..... (ii)
Subtracting (ii) from (i), we get (1 - r) Sn = (a - arn)
or Sn = a(1-rn) , for r <1.
1-r
a(rn-1)
and Sn = r -1 , for r>1.
Again, Sn = a(rn-1) , = arn- a = arn-1r - a
r -1 r -1 r -1
Since l = tn = arn-1, we get,
? Sn = lr - a It is used when the last term of G.S. is given.
r -1
Worked Out Examples
Example 1. Find the sum of the first 8 terms of the geometric series : 1+2+4+ 8 +.....
Solution: Here, 1 + 2 + 4 + 8 + ......
the first term (a) = 1
the common ratio (r) = t2 = 2 = 2 > 1.
t1 1
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 75
vedanta Excel In Opt. Mathematics - Book 10
Sum of the first eight term (S5) = ?
Now, we know that, Sn = = a(rn-1)
r -1
1(28-1)
or, s8 = 2 -1 = 256 - 1 = 255
S8 = 255
Example 2. Find the sum of the first 8 term of given G.S. 512 + 256 + 128 + ......
Solution:
Here, 512 + 256 + 128 + ..... t2 = 256 = 1 < 1
Example 3. t1 512 2
Solution: The common ratio (r) =
The first term (a) = 512
Example 4.
Solution: The sum of the first 8 term (S8) = ?
a(1- rn)
Now, we know that, Sn = 1 -r
or, S8 = 512 [1 – ( 12 )8] = 512 [1– 2156 ]
? S8 = 512 1 – 1 u2 2 –1
2 2
256 – 1
. 256
= 512 . 225556 . 2 = 1020
= 1020
Find the sum of the series: 81 + 27 + 3 + ... + 1
81
Here, first term (a) = 81
common ratio (r) = t2 = 27 = 1
t1 81 3
last term (l) = 1
81
sum of the first 8 term (S8) = ?
1 1
Now, Sn = a – lr = 81 – 81 . 3
1 -r
1 – 1
3
1
= 81 – 243 = 19683 – 1 u 2 = 19682 = 1214810
243 3 162
2
3
The sum of the first four terms of G.P. with common ratio 3 is 80. Find the
sum of the first 8 terms.
Here, common ratio (r) = 3, n = 4
Now, S4 = a(r4 –1)
r–1
a (34 – 1) a (81 – 1)
or, 80 = 3 – 1 =
or, 80 = 80a 2
2 ? a=2
Now, S8 = a(r8 –1) = 2(38 – 1)
r–1 3–1
76 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
= 2 . 6561 – 1
2
? S8= 6560
Example 5. If the third and sixth terms of 6 G.P. are 8 and 64, find the sum of the first 8
Solution: terms of the series.
Example 6. Here, the third term (t3) = 8
Solution: or, ar2 = 8 ............ (i)
The sixth term (t6) = 64
or, ar5 = 64 ......... (ii)
dividing (ii) and (i), we get
ar5 = 64
ar2 8
or, r3 = 8 = 23
? r = 2
Putting the value of r in equation (i), we get
a.22 = 8
? a = 2
Now, S8= a(r8 –1) = 2(28 – 1) = 2 . (256 – 1) = 2. 255 = 510
r–1 2–1
The sum of the first four terms is 40 and sum of first two terms is 4 in a G.P.
the common ratio is positive. Find the sum of the first 8 terms.
Here, in a given G.P.
S4 = 40 and S2 = 4
Now, Sn = a(rn –1)
r–1
a (r4 – 1)
Then, S4 = r – 1
a (r2 – 1) (r2 + 1)
or, 40 = r– 1
or, 40 = a(r + 1) (r – 1) (r2 + 1)
(r – 1)
? a(r + 1) (r2 + 1) = 40 ............ (i)
and, S2 = a(r2 –1)
r–1
or, 4 = a (r + 1) (r – 1)
r –1
or, 4 = a (r + 1)
? a(r + 1) = 4 .......... (ii)
dividing (ii) by (i), we get,
a(r + 1) 1) = 4
a(r + 1) (r2 + 40
1 1
or, r2 + 1 = 10
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 77
vedanta Excel In Opt. Mathematics - Book 10
or, 10 = r2 + 1
or, r2 = 9 = 32 ? r=±3
Taking positive square root only,
r=3
Putting the value of ‘r’ in (ii), we get,
a(3 + 1) = 4
? a = 1.
Sum of the first 8 terms (S8) = a(r8 –1) = 1(38 –1) = 6550 = 3280.
r–1 3–1 2
Example 7. Find the sum of the first 8 terms of a G.P, whose second term is 48 and fifth
Solution: term is 6.
Example 8. Here, in a given G.P.
Solution:
t2 = 48 ( tn = arn – 1 )
or, ar = 48 ............ (i)
and t5 = 6
or, ar4 = 6 ............. (ii)
Dividing (ii) and (i), we get,
( )ar4=6
48
ar 13
or, rr3==1218 = 2
?
Putting the value of r in equation (i), we get
a.21 = 48
? a = 96
Sum of the first 8 terms (S8) = a(1 –rn) [1 < r]
1 –r
96 [1 – ( 1 )8] 96 [1– 1 ]
96 2 2– 256
= 1 – 1 = 1
= 2
(256 – 1) 2 255 1
128 4
256 u2 = 96 . = 191
The sum of the first 6 terms of a G.P. is 9 times the sum of the first three
terms. Find the common ratio.
Here, in a given G.P.,
[ ]S6
= a(r6 – 1) Sn = a(rn – 1) [r > 1]
r–1 r–1
and S3 = a(r3 – 1)
r–1
Now, by the questions,
S6 = 9S3 a(r3 – 1)
r–1
or, a (r6 – 1) =9
r–1
78 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
or, (r3 + 1) (r3 – 1) = 9 (r3 – 1) ? r=2
or, r3 + 1 = 9
or, r3 = 8, r3 = 23
Therefore, the common ratio is 2.
Example 9. There are 10 varieties of birds in a zoo, the number of each variety being
double of the number of another variety. If the number of the first variety is
3, find the total number of birds of all varieties in the zoo.
Solution: Here, in given question, the varieties of birds are in G.P.
The common ratio (r) = 2
The first varieties of birds = a = 3
The total number of varieties (n) = 10
Sum of birds of all varieties (S10) = ?
a(rn – 1)
Now, S8 = r–1 = 3 (210 – 1)
2– 1
= 3 . (1024 – 1) = 3 × 1023 = 3069
Therefore, the total number of birds is 3069.
Exercise 3.6
Very Short Questions
1. If a = the first term, n = the number of term,
r = common ratio of a G.P., write the formula for sum of the first n term.
(a) When r > 1 (b) when r < 1
(c) when the last term l is given.
Short Questions
2. Find the sum of the following series :
(a) 2 + 4 + 8 + ....... upto 10 term
(b) 27 + 9 + 3 + ........ upto 10 term
(c) 1 + 1 + 1 + ....... upto 8 term
2 4
(d) 2 - 2 + 2 2 ....... upto 8 term
3. Find the sum of the following series:
(a) 3 + 6 + 12 + ........ + 384
(b) 1 + 1 + 1 + ......... + 1
3 9 243
1 1 1
(c) 1 + 2 + 4 + ....... + 256
(d) 81 + 27 + 9 + ............... + 1
243
4. Evaluate :
6 6 8
(a) ∑ 2k (b) ∑ (-1)k. 2k (c) ∑ (-1)2k.3k
k=1 k=1 k=1
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 79
vedanta Excel In Opt. Mathematics - Book 10
5. (a) If a geometric series has sum 280, common ratio 3, the last term 189 find the first
term.
(b) Find the first term of a geometric series whose last term is 1792, the common ratio
2, sum 3577.
(c) Find the first term of 6 geometric series whose sum is 728, common ratio is
3 and the last term 486.
6. (a) How many terms of G.P. must be taken in the series 128 + 64 + 32 +..... so that
the sum may be 254?
(b) How many terms a G.P. must be taken in the series 64 + 96 + 144 + 216 + ...... so
that the sum may be 2059 ?
(c) How many terms of G.P. must be taken is the series 3-6+12 - ...... so that the sum
may be - 63?
7. (a) The first term of a G.P, is 5 and the sum of the first four terms is 780, find the
common ratio.
(b) The first term of a G.P. is 3 and the sum of the first four terms is 120, find the
common ratio.
Long Questions
8. (a) The sum of the first four terms of a G.P. with common ratio 3 is 40. Find the sum
of the first ten terms of the progression.
(b) The sum of the first three terms of a G.P. with common ratio 2 is 112. Find the sum
of the first 8 terms of the progression.
9. (a) The third and the sixth terms of a G.P. are respectively 14 and 112. Find the sum
of the first 10 term terms of the progression.
(b) The 3rd and the 6th terms of a geometric series are respectively 1 and 1 . Find the
sum of the first six term of the series. 3 81
10. (a) In a G.P. the sum of three consecutive terms is 21 and their product is 64, find them.
(b) The sum of three consecutive terms in G.P. is 62 and their product is 1000, find the them.
11. (a) The sum of the first 2 terms of a G.P. is 6 and that of the first four terms is 15/2.
Find the sum of the first six term.
(b) The sum of the first 2 terms of a G.P. is 8 and that of the first 4 terms is 80. Find
the sum of the first 6 terms.
12. (a) The sum of the first 8 terms of a G.P. is 5 times the sum of the first 4 terms. Find
the common ratio
(b) The sum of the first 6 terms of a G.P. is 9 times the sum of the first 3 terms. Find
the common ratio.
13. aTrhee1s2utmeromfsthaeltfoigrsetth2etre.rmFisnodftaheG.cPo. mism89onanradtitoheansudmthoefftihrsetltaesrtm2.terms is 1152. There
14. (a) The sum of four consecutive terms of a G.S. is 30 and the arithmetic mean of the
first and the last term is 9. Find the common ratio.
(b) Three numbers are in the ratio of 1:2:3. If 2, 4 and 11 are added to them respectively,
the resulting numbers are in G.P., find the original three numbers.
80 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
15. (a) A person borrow Rs 4368 on the basis of 6 annual installments each installment
being treble of the preceding one. Find the first and the last installments.
(b) A person borrows Rs 20440 and pays it back in 9 annual installments each
installment being double of the preceding one. Find the first installment.
1. (a) (i) Sn = a(rn - 1) , r > 1 (ii) Sn = a(1 - rn) , r < 1 (iii) Sn = lr - a
r-1 1 - r r - 1
29524 255 15 2
2. (a) 2046 (b) 729 (c) 128 (d) - 1+ 2
3. (a) 765 (b) 364 (c) 511 (d) 29524
243 256 243
4. (a) 128 (b) 42 (c) 9840
5. (a) 7 (b) 7 (c) 2 6. (a) 7
(b) 7 (c) 6
7.(a) 5 1 (b) 3 40
8. (a) 29524 (b) 4080 2 81
9.(a) 3580 (b) 4
10. (a) 1, 4, 16 or 16, 4, 1 (b) 2, 10, 50 or 50, 10
11. (a) 63 (b) 728 12.(a) 2 (b) 728 13. 2, 3 14.(a) 1 or 2
8 8 2
(b)3, 6, 9, or –2, – 4, –6 15. (a) Rs 12, Rs 2916 (b) Rs 40
3.7 Natural Numbers
The numbers which are used to count objects are called natural numbers. The numbers
1,2,3..... are natural numbers. The set of natural numbers is denoted by N.
Sum of the first n natural numbers.
Let us consider the first n natural numbers.
1,2,3,......n
Then, 1 + 2 + 3 +....... +n is an A.S. with the first term = a, common difference = d = 1
Now, let Sn = 1 + 2 + 3 + ....... + n
= n [2a + (n-1)d] Where, d = 1, a = 1
2
n
= 2 [2.1 + (n-1).1]
= n (n + 1)
2
n
Sn = 2 (n + 1)
Sum of the First n odd natural Numbers. 81
The first n odd natural numbers are 1, 3, 5, ..... 2(n-1). This is an A.P.
First term = a common difference (d) = 2.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
Now, Sn = n [2a + (n-1)d]
2
n
= 2 [2.1 + (n-1).2]
= n .2.[1 + n - 1] = n2
2
⸫ Sn = n2
Example 1. Find the sum of the first 100 natural numbers. (i.e. 1 + 2 + 3 +.... + 100)
Solution: For sum of the first 100 natural numbers
n =100
Sn = n(n+1)
2
100(100 + 1) 100.(101)
or, S100 = 2 = 2 = 5050
Sum of the first n even natural numbers.
The first n even natural numbers are 2,4,6,..... 2n, which are in A.P. with common difference
(d) = 2, first term (a) = 2.
Now, Sn = n [2a + (n-1)d] = n [2.2 + (n-1).2] = n(n+1)
? 2 2
Sn = n(n + 1)
Example 2. Find the sum of the first 20 odd natural number (1 + 3 + 5 + ...... 20 term)
Solution: Here, number of terms (n) = 20
Example 3.
sum of the first 20 terms (s20) = ?
Sn = n2
or, S20 = (20)2 = 400 .
Find the sum of the first 20 even natural numbers. i.e. 2+4+6+.....20
terms.
Solution: Here, 2 + 4 + 6 + ...... 20 from number of term (n) = 20
Now, Sn = n(n + 1)
= 20 (20 + 1)
= 20. 21.= 420.
Exercise 3.7
Very Short Questions
1. Define natural numbers.
2. Give examples each of the following.
(a) First four natural number.
(b) First four even natural numbers.
(c) First four odd natural numbers.
82 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
3. Write the formula to find the sum of :
(a) The first n natural numbers.
(b) The first n odd natural numbers.
(c) The first n even natural numbers.
Short Questions
4. Find the sum of the following series:
(a) 1 + 2 + 3 + ..... + 50 (b) 1 + 2 + 3 + ...... + 100
(c) 1 + 2 + 3 + ...... + 80 (d) 1 + 2 + 3 + 4 + ...... + 1000
(e) sum of the first 40 natural numbers
5. Find the sum of the following series:
(a) 1 + 3 + 5 + ....... 10 terms (b) 1 + 3 + 5 + ...... 20 terms
(c) 1 + 3 + 5 + ....... + 99 (d) 1 + 3 + 5 + ...... + 151
(e) sum of the first 30 odd natural numbers.
6. Find the sum of the following series:
(a) 2 + 4 + 6 + ....... 20 terms (b) 2 + 4 + 6 + ........ 25 terms
(c) 2 + 4 + 6 + ....... + 100 (d) 2 + 4 + 6 + ........... + 200
7. (a) Find the sum of the sum of first 20 natural number and sum of the first 100 natural
numbers.
(b) Find the sum of the sum of first 100 natural numbers and the sum of the first 30
odd natural numbers.
(c) Find the sum of sum of the first 50 natural numbers and 30 even natural numbers.
8. (a) Find the difference of the sum of the first 20 natural numbers and the sum of the
first 15 odd natural numbers.
(b) Find the difference of the sum of the first 40 even natural numbers and the first 40
odd natural numbers.
Project Work
9. List where are the natural numbers used. Discuss in the class.
3. (a) Sn = n(n + 1) (b) Sn = n2 (c) Sn = n(n + 1) 4. (a) 1275
2 (c) 3240 (d) 500500 (e) 820
(b) 400 (c) 2500 (d) 5776 (e) 900
(b) 5050 (b) 650 (c) 2550 (d) 10100
(b) 5950 (c) 2205
5. (a) 100 (b) 40
6. (a) 420
7. (a) 5260
8. (a) -15
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 83
vedanta Excel In Opt. Mathematics - Book 10 4
Linear Programming
4.0 Review
Let us discuss the answers of the following questions :
(a) Under which conditions two real numbers ‘a’ and ‘b’ are equal ?
(b) Write any four points on the locus represented the equation x + y = 10. Draw the locus
of it.
(c) Draw graphs of 3x + 4y = 12 and 3x + 4y ≤ 12. What is differences between them?
(d) What does x = 3 represent ? Show it graphically.
(e) What does x ≥ 3 represent ? Show it graphically.
(f) Say a difference between x = 3 and x ≥ 3.
4.1 Linear Inequalities in one variable
We know that equations are mathematical statements having two sides connected by a
equality sign. In an inequality or inequation, two sides of the statements are connected by
an inequality sign. The inequality signs are >, <, ≤, ≥, ≠ 0.
Statements like ax + b > 0, ax + b < 0, ax + b ≥ 0, ax + b ≤ 0 are linear inequalities.
4x+12 > 0, 3x +15 < 0, 2x + 5 ≥ 0, 4x+18 ≤ 0, etc. are examples of linear inequality in x.
For two real numbers a and b, anyone of the following statements is true
(i) a = b (ii) a > b (iii) a < b
Two real numbers a and b are equal if their difference is zero, i.e., a - b = 0
For two real numbers a and b, if a - b = +ve, then a>b and if a – b = -ve, then a < b.
Graphical Representation of linear inequalities is one variable.
Let us consider the following linear inequities is variable x.
x > 4, x≥4, x < 4, x ≤4.
x>4 x≥4
012 34 5 012 34 5
fig (i) fig (ii)
84 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
x<4 vedanta Excel In Opt. Mathematics - Book 10
x≤4
012 34 5 012 34 5
fig (iii) fig (iv)
The graph of the inequality x > 4 is the half line or ray to the right of the point x = 4 as
shown in fig(i). Here, the point x = 4 is excluded in the solution set.
The graph of x ≥ 4 is the half line or ray to the right of x = 4 including 4 in the solution set
as shown in the fig(ii)
The graph of x < 4 is the half line or ray to the left y points x = 4 excluding the point 4 as
shown in the fig (iii).
The graph of x ≤ 4 is the half line or ray to the left of x = 4 including the point 4 as shown
in the fig (iv).
The set of points in the real number line satisfying the inequality are called the solution set.
Note :
It is to be noted that x ≥ 4 means either x = 4 or x > 4 but not both at the same time.
For example put x = 4, we get 4 ≥ 4 which is true as we take x = 4 but not 4 > 4. Also
put x = 5, we get 5 ≥ 4 which is true as we take 5 > 4 but not 5 = 4.
4.2 Linear Inequalities in two variables
A statement of inequality containing two variables x and y is known as a linear inequalities
of two variables. The following are linear inequalities of two variables.
(i) ax + by + c > 0 (ii) ax + by + c < 0
(iii) ax + by + c ≥ 0 (iv) ax + by + c ≤ 0
The set of all ordered pairs of real numbers satisfying a given inequality or inequation are
called the solutions set of the inequality.
As the set of ordered pairs of real numbers corresponds to point of a two dimensional
coordinate plane, the set of points satisfying a given inequality is known as the graph the
inequality.
Note :
Linear inequalities are also called linear inequations.
Graphical representation of linear inequalities in two variables
Inequalities like 4x + 3y - 12 ≥ 0, x + y + 3 ≤ 0, 4x + 3y > 0, 4x + 3y - 12 > 0 are called
linear inequality in two variables x and y.
Before plotting the graph of above inequalities, let us note the steps for drawing the graphs
of two variable inequalities.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 85
vedanta Excel In Opt. Mathematics - Book 10
Steps for drawing the graphs for linear inequalities in two variables
To draw the graphs of linear inequalities in two variables, the following steps are to be
followed.
(i) The inequality sign is replaced by the sign of equality to get the equation
ax + by + c = 0 which is called boundary line equation of corresponding inequality.
For strict inequalities < or >, draw the dotted line and if linear inequality containing
≤ or ≥, draw the solid line (thick line).
(ii) The eapqnoudinaYttis-oaxnxi=asxr-e+spacbeyacnt+idvecyly==c0h- omobcasyiin.ebg.etxhw-eirnisttcetearncleesipnstotahntehdafoytr-–minactoefracn–exdpact–+arbce–amybrcear=bkoetd1hoinnttehgeerXs-.
Two
axis
Alternatively, put y =0, x = 0 in ax + by + c = 0 successively to get the points in X-axis
and Y-axis respectively. Plot them in the axes and join them.
(iii) A line (dotted if the inequality contains < or > and continuous if the inequality shows
≥ or ≤ 0) is then drawn drawn through these two points. This divides the plane into
two halves.
(iv) An arbitrary point is taken in any one of the two regions to test whether it satisfies
given inequality. If it satisfies, then. The chosen region is shaded as the solution set. If
it does satisfy, the other region is shaded to denote the solution.
The origin O(0,0) can be used for the determination of the plane region of the inequality
as the test point if the corresponding boundary line does not pass through it.
Examples of Graphs of linear inequalities X' Y X
X
Example 1. (a) Draw graph of x ≥ 0 O
Write the boundary line equation x = 0 corresponding to given Y'Y
inequality x ≥ 0. Here, x = 0 represents equation of Y-axis. Since x ≥
0, its solution region is the right half plane of Y-axis including Y-axis. O
In the graph, the shaded region represents the solution of x ≥ 0.
Y'
Example 2. Draw Graph of x ≤ 0. X' Y
Write the boundary line equation x = 0. It is the equation of Y-axis.
Since x ≤ 0, the solution region is the left half plane of Y-axis including
the Y-axis. The solution region of x ≤ 0 is shaded in the graph.
Example 3. Draw Graph of y ≥ 0.
The boundary line equation of corresponding inequality is X' 0X
y = 0. It represents X-axis. Since y ≥ 0, the solution region Y'
is the upper half plane of X-axis including the X-axis. The
solution region of y ≥ 0 is shaded in the graph.
86 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
Example 4. Draw graph of y ≤ 0. Y
The boundary line equation of the corresponding inequality is OX
y = 0. It represents X-axis. Since y ≤ 0, the solution region is X'
the lower half plane of X-axis including the X-axis.
The solution region of y ≤ 0 is shaded in the graph. Y'
Example 5. Draw graphs of x ≥ 4 and x ≤ 4
Equation of boundary line in each of given inequality is x = 4. It represents a straight line
parallel to Y-axis, 4 units right of Y-axis.
The solution region of x ≥ 4 is the right half plane of the line x = 4 including the boundary
line.
The solution region of x ≤ 4 is the left half plane of the line x = 4 including the boundary
line.
YY
x≥4 x≤ 4
X' O 1 2 3 4 X X' O X
1 234
Y' Y'
Example 6. Draw graphs of y≥4 and y≤4
Equation of boundary line in each of given inequality is y = 4. It represents a straight line
parallel to X-axis, 4 units above X-axis.
The solution region of y ≥ 4 is the upper half plane of the line y = 4 including the boundary
line.
The solution region of y ≤ 4 is the lower half plane of the line y = 4 including the boundary
line.
YY
y≥4
4 4
3 3
2 2 y≤4
X' 1 X X' 1 X
O O
Y' Y'
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 87
vedanta Excel In Opt. Mathematics - Book 10
Example 7. Draw graph of 3x + 4y ≥ 12 Y
Equation of boundary line is 3x + 4y = 12. It can be written
as x + y =1 4
4 3 3
2 3x+4y≥12
1
Which is in the form of x + y = 1. Therefore, the X' -4 -3 -2 -1 O 12 3456 X
a b -1
boundary line passes through the points (4,0) and (0,3). -2
-3
-4
Take 0(0,0) as a testing point in 3x + 4y ≥ 12, we get
0 ≥ 12, which is false so the solution region of Y'
3x + 4y ≥ 12 is the half plane not containing the origin
including the boundary line.
Example 8. Draw graph of 2x + 3y ≤ 12 Y
Equation of boundary line corresponding to given X' 6 X
inequality is 2x + 3y = 12. When y=0, then x = 6. Also when 5
x = 0, then y = 4. Hence the boundary line passes through 4
the points (6,0) and (0,4). Take O(0,0) as a testing point for 3
2x + 3y ≤ 12, we get 0 ≤ 12 which is true. Hence the 2
solution region of 2x + 3y ≤ 12 is the half plane containing 1
the origin.
O 12 3456
4.3 System of linear inequalities
Y'
The combination of two or more linear inequalities having a common solution region is
called system of linear inequalities. In the system of linear inequalities, the plane region
determined by the set of inequalities is shown on the same graph. The common solution
region is shaded in the graph. Any point in the shaded region satisfied by the system of
linear inequalities. The common solution region determined by the inequalities is called
feasible region. If the solutions region is closed, it is also called convex polygonal region.
Example 9. Draw graph of 2x + y ≤ 4 and 2x - y ≤ 2 on the same graph.
Solution :
The given inequalites are, 2x + y ≤ 4 ...... (i)
and 2x - y ≤ 2.... (ii)
The corresponding boundary line equations of given in equations are:
2x + y = 4 ........... (iii)
2x - y = 2 ............ (iv) `
From boundary line equation (i) y = 4 - 2x
x20
y04
88 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
The boundary line (iii) passes through the points (2,0) and (0,4), taking a
point (0,0) for testing 2x + y ≤ 4, we get, 0 ≤ 4 which is true.
Therefore, solution of 2x + y ≤ 4 is the half plane containing the origin.
From boundary line equation (iv), y = 2x - 2.
x10
y 0 -2
Boundary lines are plotted on the same graph. Common solution region is
shaded in the graph.
Y
4
3 2x – y = 2
2
1
X' -4 -3 -2 -1 O1 2 3 4 X
-1 2x + y = 4
-2
-3
-4
Y'
Example 10. Draw graph for system of linear inequities
x + y ≤ 6, x - y ≥ -2, x ≥ 0 y≥0
Solution : The boundary line equation of given system of inequalities are given by
x + y = 6 .... (i)
x - y = -2 ...... (ii)
x = 0 ............ (iii)
y = 0 .......... (iv)
From boundary line equation, (i) x + y = 6
or, x + y = 1
6 6
x y
Which is in the from of a + b = 1
The boundary line equation (i) passes through the points (6, 0) and (0, 6).
Taking testing point 0(0, 0) for x + y ≤ 6, we get 0 ≤ 6, which is true. Hence,
the solution region of x + y ≤ 6 is the half plane containing the origin.
From boundary line equation (ii), x - y = -2
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 89
vedanta Excel In Opt. Mathematics - Book 10
or, x + y = 1
-2 2
From boundary line equation (ii) passes through the points (-2,0) and (0,2).
Taking testing point 0(0,0) for x - y ≥ -2, we get 0 ≥ -2, which is true. Hence
solution region of x-y≥-2 is the half plane containing the origin.
x = 0 and y = 0 represents Y-axis and X-axis respectively. Solution region of
x ≥ 0 and y ≥ 0 are right half plane of Y-axis and upper half plane of X-axis
respectively.
All the boundary lines are plotted and common solution region (or feasible
region) is shaded in the graph.
Y
6 x–y=2
4
2
X' -6 -4 -2 O2 4 6 X
-2 x+y=6
-4
-6
Y'
4.4 Linear Programming
In a business, the businessman is interested to get maximum profit, minimum investment,
maximum production by using limited resources. This type of problem can be solved by the
mathematical technique known as linear programming. The following are the some basic
terms used in linear programming problems.
(a) Objective function : A linear function whose maximum or minimum values or both
values are to be calculated is called objective function. For example, let x and y be
the number two types items A and B produced from a factory. On selling A and B
items there are profits of Rs. 10 and Rs. 12 per unit. Then, the profit function is
P = 10x + 12y, which is to be maximized. Then, profit function is the objective function.
(b) Constructions : The conditions or restrictions in solving a linear programming problem
which are to be satisfied by the variables of the objective function are called constraints.
In linear programming problems. The inequations are taken as constraints. In above
example, we can write condition x+y≥1000, x≥0, y≥0, which are constraints.
(c) Feasible Region (or convex polygonal Region) : A closed plane region bounded by the
intersection of finite number of boundary lines is called feasible region (or convex
polygonal region). it is the common solution region bounded by given constraints. We
may get different shapes of convex polygonal region.
90 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
(d) Feasible solution : The values of variables used is the objective function which satisfy
all the given conditions are called feasible solution .
Linear Programming : A mathematical technique of getting a maximum or minimum or
both maximum and minimum values of an objective function satisfying given constraint
is called linear programming. In linear programming problems, the objective function and
constraints must be linear.
Mathematical formulation of linear programming problems.
(i) Identify the variables and assign symbol for them.
(ii) Express the objective function as a linear function of the variables.
(iii) Identify the constraints which are to be satisfied by the variables.
Solution of linear programming problems by graphic method :
(i) Draw the graphs of each of the constraints showing their solution set by writing arrows
instead of shading.
(ii) Get the common solution of the constraints as the feasible region (convex polygonal region)
(iii) Obtain the vertices of the feasible region as the point of intersections of the boundary
lines.
(iv) Find the values of the objective function at each vertex to get maximum or minimum
values as required.
Example 11. Maximize and minimize (optimize) the objective function F = 2x + 3y
subject to constraints. x + 2y ≤ 10, 2x + y ≤ 14, x ≥ 0, y ≥ o.
Solution: The corresponding boundary line equations of the given inequalities are respectively :
x + 2y = 10 ........... (i) Y
2x + y = 14 ........... (ii) 14
13
12
x = 0 .............. (ii) 11
10
9 2x + y = 14
y = 0 ......... (iv) 8
7
6
From equation (i) y = 10 - x x + 2y = 10
2 C(0, 5)45
B(6, 2)
3
2
x02 1 A(7, 0)
y54 X' O 1 2 3 4 5 6 7 8 9 10 11 X
Y'
The boundary line (i) passes through the points (0, 5) and (2, 4), taking
0(0, 0) as a testing point and putting x = 0 and y = 0 in x + 2y ≤ 10, we get
0 ≤ 10, which is true. Hence, the solution region of x + 2y ≤ 10 is the half
plane containing the origin. From equation (ii) y = 14 - 2x.
x04
y 14 6
The boundary line (ii) passes through the points (0, 14) and (4, 6). Taking
0(0, 0) as a testing point and putting x = 0, y = 0, in 2x + y ≤ 14, we get
0 ≤ 14, which true. Hence the solution region of 2x + y ≤ 14 is the half
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 91
vedanta Excel In Opt. Mathematics - Book 10
plane containing the origin. x = 0 and y = 0 represent Y-axis and X-axis
respectively. Solution region of x ≥ 0 is the right half plane of Y-axis and that
of y ≥ 0 is upper half plane of X-axis.
All the boundary lines are plotted in the graph and common solution region
is quadrilateral OABC with vertices O(0,0), A(7,0), B(6,2) and C(0,5)
Table :
Vertices x y Value of F=2x + 3y Result
Minimum
O(0,0) 00 0
A(7,0) 70 14 Maximum
B(6,2) 62 18
C(0,5) 05 15
Hence, the maximum value of F is 18 at B(6, 2)
and the minimum value of F is 0 at O(0, 0)
Example 12. Find the maximum value of P = 3x + 2y under the constraints 2y ≥ x – 1,
x + y ≤ 4, x ≥ 0, y ≥ 0.
Solution : The corresponding boundary line equations of given inequalities are given by:
2y = x - 1 ................ (i)
x + y = 4 ................ (ii)
x = 0 ................ (iii)
y = 0 .............. (iv)
From equation (i), y = x-1
2
x15
y02
The boundary line passes through the points (1.0) and (5,2). Taking 0(0.0) as
a testing point and putting x = 0, y = 0 in 2y ≥ x-1, we get 0 ≥ -1 which is
true. The solution region of 2y ≥ x-1 is the half plane containing the origin.
From equation (ii), Y
7
y=4-x
6
5 x+y=4
x04 4 2y = x - 1
y40 3
2
The boundary line (ii) passes through the 1
points (0, 4) and (4, 0). Taking 0(0, 0) as a
X' O 1234567 X
testing point and putting x = 0 and y = 0
in x + y ≤ 4, we get 0 ≤ 4, which is true. Y'
The solution region of x + y ≤4 is the half
plane containing the origin. x = 0 and y = 0 are equation of Y-axis and X - axis
respectively. Solution region of x ≥ 0 is the right half plane of Y-axis and that
92 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
y ≥ 0 is the upper half plane if X-axis.
All the boundary line are plotted in the graph and common solution is convex
polygon OABC with vertices 0(0,0), A(1,0), B(3,1) and C(0,4) which is shaded
in graph.
Table :
Vertices xy P=3x + 2y Result
Minimum
O(0,0) 00 0
A(1,0) 10 3 Maximum
B(3,1) 31 11
C(0,4) 04 8
Hence, the maximum value of P is 11 at B(3,1).
Example 13. In the given graph shaded region represented by the ∆PQR gives the feasible
region. Write down the inequalities corresponding to the given boundary
lines.
Solution : The vertices of the feasible region ∆PQR are P(5.0), Q(6,7), and R (0,6). The
boundary lines are PQ, QR, and RP.
For the boundary line PQ. Y
It passes through the points P(5,0) and Q(6,7). 8
So, its equation is given by, 7 R(0, 6) Q(6, 7)
6
0y1==76yx2--2,-05-xy(11x(-x5-)x1) 5
y - 4
y -
3
or, 2
or, y = 7 (x-5) X' 1 X
P(5, 0)
O 1234567
or, y = 7x - 35
or, 7x - y = 35 Y'
The half plane including the boundary line PQ contains the origin. So, the
corresponding inequality of boundary line PQ is 7x - y ≤ 35.
For the boundary line QR.
It passes through the points Q(6, 7) and R(0,6). Its equation is given by,
y-7 = 6-7 (x - 6)
0-6
1
or, y - 7 = -6 (x – 6)
or, 6y - 42 = – x + 6
or, x + 6y = 48
The half plane including the boundary line QR contains the origin. So, the
corresponding inequality of boundary line QR,
x + 6y ≤ 48
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 93
vedanta Excel In Opt. Mathematics - Book 10
For boundary line PR.
It passes through the points P(5,0) and R(0,6). If x-intercept = 5
and y-intercept = 6, so, its equation is
x + y =1
5 6
6x+5y
or, 30 = 1
or, 6x + 5y = 30
The half plane including the boundary line PR does not contain the origin.
The corresponding inequity of boundary line PR is:
or, 6x + 5y ≥ 30.
Hence, the required inequalities are
7x – y ≤35, x + 6y ≤ 48 and 6x + 5y ≥ 30.
Exercise 4
Very Short Questions
1. (a) Write the boundary line equation of 4x + 3y ≤ 12.
(b) Define a feasible region.
(c) Define system of inequalities.
(d) Define linear programming.
2. Define the following terms relating linear programming:
(a) Objective function (b) Constraints
(c) Feasible solution (d) Convex polygonal region
Short Questions
3. Draw graph of the following inequities:
(a) x ≥ 0 (b) x ≤ 0 (c) y ≥ 0 (d) y ≤ 0
(e) x ≥ 7 (f) y ≤ 5 (g) 4x + 3y ≥ 12 (h) x + y ≤ 6
(i) y ≤ 2x + 1 (j) y ≤ 2x + 4 (k) 4y - 5x ≥ 20
4. Graph the following system of inequalities and find the vertices of convex polygon if
they exist.
(a) x + y ≥ 3 and 2x - y ≤ 4 (b) x + 2y ≤ 12 and x ≥ 6, y ≥ 0
(c) x + y + 2 ≥ 0 and y ≤ 2x + 4, y ≤ 4 - 4x
(d) x + y ≤ 12 and x ≥ 0, y ≥ 0
Long Questions
Solve graphically:
5. (a) Maximize Z=2x+3y subject to the constraints 2x+y ≤ 14, x + 2y ≤ 10, x ≥ 0.
(b) Maximize F = x + 2y subject to the constraints x + y ≥ 2, 2x - y ≤ 4, y ≤ 2.
94 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
(c) Maximize Z=x+y subject to constraints x+y≥3, 2x+3y≤18, x≤6, x≥0, y≥0.
(d) Maximize P = 6x + 5y subject to constraints x + y ≤ 6, x - y ≥ -2, x ≥ 0, y ≥ 0.
(e) Find the maximum value of the objective function F = 6x + 15y subject to
4x – 3y ≥ -15, x ≤ 3, y ≥ 5.
6. (a) Minimize F=2x+y subject to the constraints 2x+y≤20, 2x+3y≤24, x≥0, y≥0.
(b) Minimize P=4x-y under the constraints. 2x+3y≥6, 2x-3y≤6, 2x-3y≤6, y≤2.
(c) Minimize C = 20x + 30y subject to the constraints 3x + 5y ≥ 45, 2x + y ≥ 20,
(d) Minimize Z=12x+3y subject to the constraints 4x+y≤12, x+2y≥10, x≥0, y≥ 0
(e) Find the minimum value of Z = 10x + 12y under the given constraints:
x + 2y ≤ 12, 3x + 2y ≤ 24, x ≥ 0, y ≥ 0.
7. (a) Find the extreme values of the function F = 10x + 15y under the convex
polygon determined by the system of inequalities x + 2y ≤ 20, x + y ≤ 16,
x ≥ 0, y ≥ 0,
(b) Find the extreme value of the function F defined by F = 7x + 6y subject to
constraints. 4x - 9y ≤ 36, x ≥ 9, x ≥ 9, y ≤ 4.
(c) Maximize and minimize the objective function F = 6x + 5y subject to the
constraints x + y ≤ 6, x - y ≥ -2, x ≥ 0, y ≥ 2.
(d) Maximize and minimize function F = x + 2y subject to the constraints
2x + y ≤ 14, x+2y ≤ 10, x ≥ 0, y ≥ 0.
(e) Find the maximum and minimum value of the objective function
F = 16x - 2y + 40 subject to 3x + 5y ≤ 24, 0 ≤ x ≤ 7, 0 ≤ y ≤ 4
8. (a) In the given graph, the coordinates of 8 Y
the points are A(2, 0), B(1, 2), C(0, 3),
D(0, 5) and E(5, 0). OABC is the feasible 7
region. Find all the inequalities and
minimize P = 6x + 10y + 20. 6
(b) In the given diagram, ∆ABC is the feasible 5 D(0, 5)
region. AB is parallel to the Y-axis and the
coordinates of B are (5,10), one of constraint 4
is x + y ≥ 5, Find the remaining constraints 3 C(0, 3)
and minimize C = 3x + 5y.
2
X' 1 A(2, 0) E(5, 0) X
O 1234567
Y'
11 Y
10 B(5, 10)
9
8
7 C(0, 5)
6
5
4
3
2
X' 1
A(5, 0) X
O 123456789
Y'
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 95
vedanta Excel In Opt. Mathematics - Book 10 8 Y
(c) In the given diagram OABC is the feasible
region. The coordinates of B and C are (2,4) 7 (0, 6)
and (0,2) respectively. If one of the constraint 6
is x + y ≤ 6, find all other constraints and
find the maximum value of Z = 6x + 5y. 5
(d) Write down the inequalities to represent the 4 B(2, 4)
feasible region shaded in the graph.
C(0, 232)
X'
X' 1 A(6, 0) X
(e) In the given graph, the coordinates of A, B, O 1234567 X
and C are (2,0), (4,1), and (0,5) respectively.
OABC is the feasible region. Find the four 11 Y Y'
inequalities of feasible region.
10
9
8
7 C(0, 7)
6
5
A 4 E(4, 3)
3
2
1 B
X
O 1 2 3 4 5 6 7 8 9 10
Y' Y B(4, 1)
X' 8
7
6
5 C(0, 5)
4
3
2
1
O 1 A2(23, 0)4 5 6 7
Y'
4. (a) Convex polygon does not exists. (b) (6, 0), (12, 0), (6, 3)
(c) (-2, 0), (2, -4), (0, 4) (d) (0, 0), (12, 0), (0, 12)
5. (a) 18 at (6, 2) (b) 7 at (3, 2) (c) 9 at (9, 0) (d) 36 at (6, 0)
(e) 153 at (3, 9) 6.(a) 0 at (0, 0) (b) -2 at (0, 2)
(c) 285.71 at 55 , 30 (d) 15 at (0, 5) (e) 0 at (0, 0)
7 7
7. (a) Max. 180 at (12, 4), Min. 0 at (0, 0) (b) Max. 150 at (18, 4), Min. 63 at (9, 0)
(c) Max. 36 at (6, 0), Min. 0 at (0, 0) (d) Max. 10 at (6, 2), Min. 0 at (0, 0)
(e) Max. 152 at (7, 0) and Min. 0 at (0, 0)
8. (a) 5x + 2y ≤ 10, 3x + 5y ≤ 15, x ≥ 0, y ≥ 0, Min. value 20 at (0, 0)
(b) x + y ≥ 5, 2x - y ≥ 0, x ≥ 0, y ≥ 0, Min. 15 at (5, 0)
(c) y - x ≤ 2, x ≥ 0, y ≥ 0, Max. 36 at (6, 0)
(d) x + 2y ≤ 10, x + y ≤ 7, x ≥ 0, y ≥ 0
(e) x + y ≤ 5, x - 2y ≤ 2, x ≥ 0, y ≥ 0
96 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10 5
Quadratic Equations and
Graphs
5.0 Quadratic Equations and Graphs
Introduction :
Graphical presentation of a function is one of effective ways of studying functions. Graphs
give a bird’s eye view of functions. But graphs have their importance and limitations.
Let y = f(x) be a function. Here x is taken as the independent variable and y as the dependent
variable. Values of x are taken as the x-coordinates the corresponding value of y as the
y-coordinates. The points obtained are plotted in a graph and joined them to get the graph
of a function.
5.1 Graph of linear a Function
The general form of a linear function is y = ax + b, a ≠ 0, a and b are constants.
The equations of the first degree in x and y are called linear Y
equations. The linear equations give straight lines when plotted 8
in a graph. 7
6
Let us consider a linear function, y = 2x + 5. Then some points 5
are taken as shown in the table below : 4
3
x -2 0 1 2
1
y 1 57 X' X
-3 -2 -1O 1 2 3
Plotting the points (-2, 1) (0, 5), (1, 7) and joining them, a straight Y'
line is obtained as shown in the graph.
5.2 Graph of a Quadratic Function
A function of the second degree in variable x represented by f(x) = ax2 + bx + c, a ≠ 0, where
a, b and c are constants is called a quadratic function. For an example, y =f(x) = x2 - 6x + 5
is a quadratic function of x. It gives a special type of curve which is called a parabola.
To draw a graph of above function, let us take values of x and the corresponding values of y
in the table given below:
x -1 0 2 3 4 5 67
y 12 5 -3 -4 -3 0 5 12
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 97
vedanta Excel In Opt. Mathematics - Book 10
Y M
12
11
10
9
8
7
6
C 5
4
3
2
1A B
X' O 1234567 X
P
Y'
Plotting these points on graph and joining them successively, we get a curve called parabola.
Let us define some basic terms used in parabola.
(a) Vertex : The turning point of parabola is called its vertex. In the above curve of parabola
the point P is vertex of the parabola.
(b) Line of symmetry : The vertical line passing through the vertex of parabola and parallel
to Y-axis is called the line of symmetry. In the above graph of parabola, the line PM is
called the line of symmetry of the parabola. The line of symmetry divides the parabola
into two equal parts.
(c) Points on the axes : The point of intersection of the parabola and the coordinate axes of
graph are the points on the axes. In the above graph, the point A(1,0) and B(5,0) are on
the X-axis and C(0,5) is point on the Y-axis.
Note :
(a) If quadratic equation is x2 - 6x +5 = 0, the points on the X-axis gives the solution of
the quadratic equation. In the above graph, A and B give the solution of the quadratic
equation x2 - 6x + 5 = 0.
(b) In the quadratic function y = ax2 + bx + c, a is called the leading coefficient. If a>0,
the parabola opens upward, and the vertex is the point with the minimum value of y
on the graph. If a < 0, the parabola opens downward, and the vertex is the point with
the maximum value of y on the graph.
5.3 Graph of a Quadratic Function of y = ax2 form
Let us draw a graphs for y = ax2, for different values of a. Y
(a) Put a = 1, we get y = x2 9
8
7
x -3 -2 -1 0 1 2 3 6
y94101 4 9 5
4
3
Let us plot all these points on the table on a graph paper and 2 y = x2
join them successively. We get a graph of parabola as shown 1
in the figure. X' -4 -3 -2 -1O 1 2 3 4 X
Y'
98 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
vedanta Excel In Opt. Mathematics - Book 10
(b) Put a = 2, we get y = 2x2
x -3 -2 -1 0 1 2 3
y 18 8 2 0 2 8 18
Let us plot all the above points on the graph and join them successively, we get a
parabola as shown in the graph.
Y
18
16
14
12
10 y = 2x2
8
6
4
2
X' O 2 4 6 8 X
Y' Y
9
(c) Put a = 1 , we get y = 1 x2 8
2 2 7
6 2
5
x -4 -2 -1 0 1 2 4 4 x
3
2 1
1 2
O 1234
1 1 =
2 2
y 8 2 0 2 8 y
All there points of the table are plotted in graph and joined X' X
successively, the curve of a parabola is obtained.
Y'
Y
(d) Put a = -1, we get y = - x2 X' O 1234 X
x -3 -2 -1 0 1 2 3 -1
y -9 -4 -1 0 -1 -4 -9 -2
-3 y = -x2
-4
-5
Plotting these points on the graph and joining successively, -6
the curve of a parabola is obtained as shown in the given -7
figure. -8
-9
Y'
Y
(e) Put a = -2, we get y = -2x2 23 X' O 2 4 6 8 X
-8 -18 -2
x -3 -2 -1 0 1 -4
y -18 -8 -2 0 -2 -6
-8
Plotting there points, the parabola obtained is
as shown in the figure. -10 y = -2x2
-12
-14
-16
-18
Y'
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 99
vedanta Excel In Opt. Mathematics - Book 10
(f) Put a = - 1 , we get y = - 1 x2 Y
2 2
O 1234
x -4 -2 1 0 1 2 4 X' -1 X
-2
y -8 -2 - 1 0 - 1 -2 -8 -3
2 2 -4
-5 y
-6
-7 =
-8
From this table, plotting points, we get the curve of -9 -
parabola as shown in the figure.
1
2
x2
Now, let as plot above six curves of parabola in the same Y'
graph paper and compare the curves, discuss about them, and draw conclusions.
Y98y=2xy2 = x2
7 y
= 21x2
6
5
4
3
2
1
X' -5 -4 -3 -2 -1 O-11 2 3 4 5 X
-2
-3
-4
-5
-6
-7 y = 21x2
y=2x2
Y' y = x2
We have studied the different curves as shown in the graphs. We draw the following
conclusions for the curves of y = ax2
1. If a > 0, the parabola opens upwards.
2. If a < 0, the parabola opens downwards.
3. If a is positively or negatively decreased, the mouth of the parabola becomes narrower.
4. If a is positively or negatively increased, the mouth of the parabola becomes wider.
5. The vertex of parabola is at the origin.
6. Y-axis or x = 0 is the line of symmetry of the parabola.
5.4 Graph of Quadratic Function of the form y = ax2 + bx + c
General form of a quadratic function is y = ax2 + bx + c, a≠0.
For different values of a, b, c, we get different curves known as parabola. To draw the graph
of a quadratic function, we have to find the vertex of the curve.
Here, y = ax2 + bx + c
100 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Vedanta Opt. Maths Book 10 Final (2078)
Discover the best professional documents and content resources in AnyFlip Document Base.
Search