Prime Optional Mathematics 10

Again,
or, 1tn4 = = a 8 +4 ( +n (–1 15) –d 1)d
or, –70 = 14d
\ d = –5
\ common difference is –5.

6. If 12th term and 20th term of an AP are 46 and 78 respectively, find the sum of the first 25

terms.

Solution :
In an AP,
1220tthh tteerrmm ((tt2120)) == 7468


We have,
4ttn162 = == a aa + ++ ( (n11 1–2d 1–) 1d)d
or,
or,
or, a = 46 – 11d ........................................... (i)

Again, [From eqn. (i)]
or, 7t280 == a4 6+ –(2 101 –d 1+) 1d9d
or, 32 = 8d
\ d = 4

Substituting the value of ‘d' in eqn. (i)
a = 46 – 11 × 4
= 46 × 44
= 2

Again, sum of the 25 terms is,
Sn = n [2a + (n – 1)d]
2

or, S25 = 25 [2 × 2 + (25 – 1)4]
2

or, S25 = 25 × 100
2

\ S25 = 1250

46 PRIME Opt. Maths Book - X

7. If sum of the first 5th terms and first 15 terms of an AP are 110 and –30 respectively find the

sum of the first 20 terms.

Solution :
In an AP,
SSS512 50= == 1 ?–1300
We have,
n
Sn = 2 [2a + (n – 1)d]

or, S5 = 5 [2a + (5 – 1)d]
2

or, 110 × 2 = 10a + 22d

or, a = 22 – 2d ................. (i)

Again,

S15 = 15 [2a + (15 – 1)d]
2

or, 30 = 15(a + 7d)

or, 2 = 22 – 2d + 7d

or, –20 = 5d
\ d = – 4

Substituting the value of 'd' in equation (i)
a = 22 – 2 × (–4)

\ a = 30

Again, n
2
S20 = [2a + (n – 1)d]

= 20 [2 × 30 + 19(–4)]
2

= 10 × (–16)
= – 160
\ S20 = – 160

8. The 9th term of an AP is 5 times the second term and sum of the first 12 terms is 600. Find

the sum of the first 20 terms.

Solution :
St91 2= = 5 6t200
We have,
Totnar ,=k aian +g+ , ( (t9n9 –=– 151)t)d2d = 5[a + (2 – 1)d]


or, a + 8d = 5a + 5d
or, 3d = 4a
4a
\ d = 3 .............. (i)

PRIME Opt. Maths Book - X 47

Again, n
2
Sn = [2a + (n – 1)d]

or, S12 = 12 [2a + (12 – 1)( 4a )]
2 3

or, 600 = 2a + 44a
6 3

or, 100 = 50a
3

\ a = 6

Substituting the value of ‘a' in equation. (i)
4 ×6
d = 3 = 8

Again,

Sum of 1st 20 terms is,
20
S20 = 2 [2a + (20 – 1)d]

= 10[2 × 6 + 18 × 8]
= 10 × 164
= 1640

9. The sum of the three terms in AP is 15 and their product is 80. Find the terms.
Solution :

Let, the three terms in AP be a – d, a and a + d.
1st condition;
sum of the terms = 15
or, a – d + a + a + d = 15
or, 3a = 15
\ a = 5

2nd Condition;
product of them = 80
or, (a – d) a(a + d) = 80
or, 5(5 – d)(5 + d) = 80
or, 25 – d2 = 16

or, d2 = 9
\ d = ±3

Then,
a – d = 5 – (±3) = 2 or 8
a = 5
a + d = 5 + (±3) = 8 or 2
\ The required three terms are 2, 5, 8 or 8, 5, 2.

48 PRIME Opt. Maths Book - X

10. The sum of the first 4 terms of a AP is 38 whose last term is 44 and the sum of the last 4

terms is 158. Find the number of terms of the sequence.

Solution :
In and AP,
Sum of first 4 terms (s) = 38
Sum of last 4 terms= 158
Last term (tn) = 44
We have,

Sum of last four terms = 158
or, (4t n×) 4+4 ( t–n 6–d d =) +1 5(t8n – 2d) + (tn – 3d) = 158
or,
or 176 – 158 = 6d
or, d = 3

Again, n
2
Sn = [2a + (n – 1)d]

or, S4 = 4 [2a + (4 – 1)3]
2

or, 38 = 2[2a + 9]
or, a = 5

Again,
or, 4tn4 = = a 5 + + ( n(n – –1 1)d)3
or, 13 = n – 1
\ n = 14 terms

\ There are 14 terms in the sequence.

Remember :
• The three terms in AP : a – d, a and a + d.
• The four terms in AP : a – 3d, a – d, a + d and a + 3d.
• The five terms in AP : a – 2d, a – d, a, a + d and a + 2d.

PRIME Opt. Maths Book - X 49

Exercise 1.3 'B'

1. Answer the following questions.
i) Write down the formula of sum of first n terms of the arithmetic series.
ii) Are the terms a – d, a and a + d in AP?
iii) Write down the formula of sum of first 'n' terms of the arithmetic series having last

term 'l'.
iv) Write down the sequence having first term 3 and common difference 4.
v) Are the term of 4, 7, 10, 13, ..., ..., ... in AP?

2. Write down the following in expanded form and find the sum of the terms.
i) /4 (3n–2) /5 iii) /6 (5n–3)
ii)
(–1)n (2n + 3)
n=1 n=1 n=1

iv) /5

v) (n2 –3n)

n=1

3. Find the sum of the following terms of the series.
i) 2 + 8 + 14 + 20 + ... + ... + ... + 15 terms.
ii) 100 + 96 + 92 + ... + ... + ... + 20 terms.
iii) 3 – 7 – 17 – 27 – ... – ... – ... – 12 terms.
iv) 7 + 13 + 19 + ... + ...+ ... + 79.
v) 80 + 72 + 64 + ... + ...+ ... + 8

4. i) If first term is 15 and common difference of the series is 6, find the sum of the first
12 terms.
ii) If first term of an AP having sum of the first 15 terms 525 is 7, what will be its
common difference?
iii) HIf odw = m10a,n Syn t=e r1m1s7 0ar oef tthheer efi rinst t 1h5e tseerrmiess, 1fi3n d+ 2th0e + f i2rs7t +te 3r4m + o .f.. t+h .e.. +se ..q. u, wenhceer.e sum of
iv) terms is 1590 ?

v) The sum of the first ‘n' terms of the series 124 + 112 + 100 + 88 + ... + ...+ ... is 600,
find the value of ‘n'.

5. i) The 7th and 13th terms of an AP are 150 and 60 respectively. Find the sum of the first
18 terms.

ii) The 12th and 21st terms of an AP are 54 and 0 respectively. Find the sum of the first
35 terms.

iii) If 11th term of an AP is 4 times the third term and sum of the first 17 terms is 884.
Find the last term.

iv) The 15th term of an AP is 101 and sum of the first 6 terms is 123. Find the sequence.
v) The 11th term of an arithmetic series is 20 and sum of the first 16 terms is 480. Find

the series.

6. i) The sum of the first 6 terms and first 17 terms of an AP are 150 and 1173 respectively.
Find the 20th term of the series.

50 PRIME Opt. Maths Book - X

ii) The sum of the first 11 terms and first 19 terms of an AP are 660 and 1748
respectively. Find the sum of the first 30 terms.

iii) The sum of the first 12 terms of an AP is 480 and sum of the next 12 terms is 1344.
Find the 25th term of the sequence.

iv) Monthly fees of a private school of grade X is decided Rs. 5,000 where fees increased
every month by Rs. 500. Find the amount paid by a person altogether when his child
will be at the end of grade X.

v) Monthly salary of Pranav in a private company is Rs. 20,000 where his salary
increases every month by 10%. If the amount received by him after some months
altogether at all is Rs. 3,30,000, find the number of months for the amount.

7. i) Find the three terms in AP whose sum is 18 where the greatest is 5 times the least.
ii) Find the three terms in AS whose sum is 21. Where sum of their squares is 179.
iii) Find th three terms in AP whose sum is 24 and product of their extremes is 39.
iv) The first and last term of an AP are 2 and 29 respectively and sum of the terms is

155. Find the number of terms and common difference. If 5 more terms are added
to the series, what will be the last term and sum of the terms?
v) A man have to be paid a loan taking for agricultural purpose including principal and
interest is Rs. 29,000 in monthly installment and complete the loan in 20 months
where Rs. 100 more have to paid in each installment. Find the amount have to be
paid in first installment.

8. PRIME more creative questions:
i. The sum of first 6 terms of an AP is 480 and sum of next 6 terms is 768. For how

many terms the sum will be 1740?
ii. If pth term of an AP is ‘q' and qth term is p, prove that the nth term is p + q – n.
iii. Divide 60 into 4 parts which are in AP where greatest is 4 times the least.
iv. Find the sum of the terms between 100 to 200 which are, divisible by 6.
v. Sum of the first 5 terms of an AP is 70, sum of last 5 terms is 310 and the last term

is 70. Find the number of terms of the sequence.

Answer

1. Show to your teacher.

2. i) 22 ii) – 9 iii) 87 iv) 9 v) 10
v) 440
3. i) 660 ii) 1240 iii) – 624 iv) 559 v) 15

4. i) 576 ii) 4 iii) 8 iv) 20 v) 11

5. i) 2025 ii) 630 iii)

iv) 3, 10, 17, 24, ..., ..., ... v) 60 + 56 + 52 + 48 + ... + ...+ ...

6. i) 157 ii) 4080 iii) 151 iv) 93,000

7. i) 2, 6, 10 ii) 3, 7, 11 or 11, 7, 3

iii) 3, 8, 13 or 13, 8, 3 iv) 10, 3, 44, 345 v) 500

8. i) 15 iii) 6, 12, 18, 24 iv) 2550 v) 17

PRIME Opt. Maths Book - X 51

Geometric Sequence (Progression)

A sequence is called an geometric sequence if ratio of any two consecutive terms remain same.

The sequence in which the terms are written under the rule of

multiplication with a constant number is called geometric sequence.

The constant number so used in the sequence is called common ratio.
t2
i.e. r = t1

Examples :
i. 2, 6, 18, 54, ..., ..., ... each successive terms are increased by the rule of multiplication of 3.
ii. 80, 40, 20, 10, ..., ..., ... each successive terms are decreased by the rule of multiplication of 2.
iii. a, ar, ar2, ar3, ..., ..., ... each successive terms are multiplied by a constant ‘r'.
In the examples explained above, the successive terms of the sequence are written under the
rule of multiplication by (... × 3), (... × 2) and (... × r) respectively and so all of them are the
geometric sequence.

If the terms are added (+) as below, they are called series.
2 + 6 + 18 + 54 + ... + ...+ ...
80 + 40 + 20 + 10 + ... + ...+ ...
a + ar + ar2 + ar3 + ... + ...+ ...

nth term of geometric sequence :

Let us consider ‘a' is taken as the first term and ‘r' is taken as the common ratio of a geometric

sequence. The successive terms can be written as below,
14253Intsnrthhdtd t tttttheeeeeirrrrrsmmmmm w ====a= y ttttt,1453 2 n =====t h aaaaat rrrer=432 r = a===m a ×aaaro rrr1 f 2 543 –t = ––– h1 11 1aers0e =q aure n1 –c 1e can be written as,
tn = arn – 1
Where, common ration (r) =


t2
t1

Geometric Mean

Let us consider GM is the middle term of the two terms ‘a' and ‘b' of a geometric sequence.
i.e. a, GM and b are in GP.

Here, t2 t3
t1 t2
Common ration = =

or, GM = b
a GM

or, GM2 = ab
or, GM = ab
\ GM = ab (It is the geometric mean between any two terms a and b of a GP)

52 PRIME Opt. Maths Book - X

• For more than one geometric mean m1, m2, m3, ..., ..., ... mn within any two terms ‘a' and ‘b'.
The sequence so formed becomes a, m1, m2, m3, ..., ..., ... mn, b.
Here,

First term = a
Common ratio = r
Last term = b
Number of terms = n + 2 = n'

We have,
or, tbn == aarr nn '+ – 2 1 – 1
b
or, a = r n + 1

b 1
a
or, r = a kn + 1

Then, the geometric means will be,

b 1
a
m1 = t2 = ar = a. a kn + 1

b 2
a
m2 = t3 = ar2 = a. a kn + 1

b 3
a
m3 = t4 = ar3 = a. a kn + 1

Similarly as above, we get
n
b
mn = tn + 1 = arn = a. a a kn + 1

Difference between arithmetic and geometric sequence

Arithmetic sequence Geometric sequence

1. It is the sequence where the terms are in 1. It is the sequence where the terms are in

increasing or in decreasing order with a increasing or in decreasing order with a

constant number additionally. constant number multiplicatively.
(d = t2 – t1) t2
(r = t1 )

2. It's nth term is, 2. It's nth term is,
tn = a + (n – 1)d tn = a.rn–1

3. AM between any two numbers is, 3. GM between any two numbers is,
a + b GM = a.b
AM = 2

4. Sum of the first n terms is, 4. Sum of the first n terms is,
n a (rn – 1)
Sn = 2 [2a + (n – 1)d] Sn = r –1

5. Number of terms written in AP, 5. Number of terms written in GP,
3 terms = a – d, a & a + d
4 terms = a – 3d, a – d, a + d & a + 3d 3 terms = a , a & ar
5 terms = a – 2d, a – d, a, a + d & a + 2d r

4 terms = a a ar & ar3
r3 , r,

5 terms = a , a , a, ar & ar2
r2 r

PRIME Opt. Maths Book - X 53

Worked out Examples

1. Identify the type of given sequence 3, 12, 48, ..., ..., ...,

Solution :

The given sequence is, 3, 12, 48, ..., ..., ...
t2 t3
Here, Common ration (r) = t1 = t2

or, 12 = 48
3 12

or, 4 = 4
\ It is the geometric sequence.

Note : In (A), ab is the geometric mean between 'a' and 'b' if both 'a' and 'b' are +ve and
– ab is the geometric mean between 'a' and 'b' if both 'a' and 'b' are -ve.

2. Is 4374 a term of the sequence 2, 6, 18, 54, ..., ..., ..., ?

Solution :
The given sequence is, 2, 6, 18, 54, ..., ..., ...

Here, t2 t3
t1 t2
Common ration (r) = =

= 6 = 18
2 6

= 3 = 3 (It is in GP)
Fnithr stet rtmer m(t n()a =) =4 3274

Then,
or, 4tn3 =7 a4r =n- 12 × 3n – 1
or, 2187 = 3n – 1
or, 37 = 3n – 1
or, n – 1 = 7
\ n = 8
Hence, 4374 is the 8th term

3. If 7th term and common ratio of a GP are 640 and 2 respectively. Find the first term of the

sequence.

Solution :
Common ration (r) = 2
F7ithr stet rtmer m(t 7()a =) =6 4?0

We have,
6tt7n4 ==0 aa =(r 2na –) ×17 – 6 14
or,
or,
\ a = 10

54 PRIME Opt. Maths Book - X

4. If 3rd term and 8th term of a GP are 512 and 16 respectively. Find the 12th term.

Solution :

83trhd tteerrmm ((tt83)) == 15612
We have,
2ttn31 ==5 aa =rr 3na ––r 112 ............. (i)
or,
or,

Again,
or, t186 = = a ar8r 7– 1........... (ii)

Dividing equation (i) by (ii)
512 = ar2
16 ar7

or, 32 = 1
r5

or, r5 = a 1 5
2
k

\ r = 1
2

Substituting the value of ‘r' in equation (i)
1 2
512 = a a 2
k

or, a = 512 × 4
\ a = 2048

Again,
12th term is,
t12 = ar12 – 1
1 11
= 2048 × a 2
k

= 2048 × 1
2048

= 1

5. Find the geometric mean between the terms 1 and 64.
4
Solution :

First term (a) = 1
4

Last term (b) = 64

Then, Geometric mean,

GM = ab

= 1
4 × 6416

= 16
= 4

PRIME Opt. Maths Book - X 55

6. Insert 4 geometric means between 1 and 64.
16
Solution :

First term (a) = 1
16

Last term (b) = 64
No. of term (n) = 4 + 2 = 6

Then,

We have,

b 1
a
Common ration (r) = a kn–1

64 1
1
= e 6–1

o

16

= 1

^210h5

= 4

Then, the four geometric means are, Alternative method

m1 = ar = 1 × 4 = 1 tn = arn – 1
16 4 1
or, 64 = 16 × r6 – 1
m2 = ar2 = 1 × 42 = 1
16 or, 64 × 16 = r5
or 210 = r5
m3 = ar3 = 1 × 43 = 4 \ r = 4
16

m4 = ar4 = 1 × 44 = 16
16

\ The four GMs are 1 , 1, 4 and 16.
4

7. How many GMs are there between 6 and 1536 where the ratio of first mean and last mean

is 1:64?

Solution :
Let, number of GMs be ‘x'.
First term (a) = 6
NLaos ot ft etermrm (st n()n =) 1=5 x3 +6 2

Ratio of 1st mean and last mean
i.e. m1 : mx = 1 : 64

We have, Dividing equation (ii) by (i)

m1 = 1 rx+1 = 256 4
mx 64 rx–1 64
or,
ar 1
or, arx = 64 or, r2 = 22
\ r = 2
or, rx – 1 = 64 .............. (i)

Again, Again, Putting the value of ‘r' in equation (i)
or, 1tn5 =3 a6r =n –a 1 × rx + 2 – 1 2x – 1 = 26
or, x – 1 = 6
or, 1536 = rx + 1 \ x = 7
6 \ There are 7 GMs.

or, rx + 1 = 256 ........... (ii)

56 PRIME Opt. Maths Book - X

Exercise 1.3 'C'

1. Answer the following questions.
i) What is geometric sequence? Write down with an example.
ii) Prove that GM between 'a' and 'b' is ab.
iii) Write down the calculating formula of common ratio while first term 'a' and last

iv) term 'b' are given.
v) APrroev 3e, t6h, a1t2 t,n 2 =4 a, .r..n,– 1...i,n ...a inG PG.P?


2. Which of the followings are the geometric sequence? Write down with reasons.
1 1
i) 2, 6, 18, 54 ii) 1, 4, 9, 16 iii) 4, –1, 4 , – 16
5 v) a, ar2, ar4, ar6
iv) 80, 20, 5, 4

3. i) Find the next 3 terms of the sequence – 1 , – 1 , – 1, – 4,
16 4

ii) If first term and common ratio of a GP are 6 and 2 respectively, what will be its 8th term?
5
iii) If first term and fifth term of a GP are 160 and 8 respectively. Find the common

ratio of the sequence.

iv) If nth term of a GP having 10 terms and common ratio 1 is 4 find the first term.
1 1 2
64 4
v) If tn = , r = – and a = 4, find the numbers of terms of GP.

4. i) How many terms are there in the sequence 3, 6, 12, ..., ..., ... 1536?

ii) Find the number of terms of the sequence 1 , 1 , 1 ..., ..., ... 243.
9 3

iii) Find the 9th term of the series 4 – 1 + 1 – 1 ... + ... – ... + .
4 16

iv) Is 3072 a term of the sequence 12 , 24, 48 ..., ..., ...

v) Which term of the sequence 1 , 1 , 1 , – 1, ..., ..., ... is 128?
8 –4 2

5. i) Second term and 6thterm of a GP are 10 and 160 respectively. Find the 9th term of the sequence.
1 4 64
ii) If 3rd term and 7th term of a GP are 12 and 3 respectively, which term will be 3 ?

iii) The 4th term and 8th term of a GP are 1 and 9 respectively. Find the sequence.
9

iv) The 3rd term and 8th term of a GP are 640 and 20 respectively. Find the series.
v) If fifth term of a GP is 48 and its 11th term is 3072, find the 15th term of the sequence.

6. Find the geometric mean between the following pairs of terms.
i) 4 and 100 ii) –12 and – 108 iii) ar2 and ar6
1
iv) (x + y)2 and (x – y)2 v) 20 and 500.

7. i) If geometric mean between x and 100 is 20, find the value of ‘x'.
1
ii) Insert 4 GMS between 4 and 8.

PRIME Opt. Maths Book - X 57

iii) Insert 5 GMS between 81 and 1 .
2 18

iv) If 320 p, q, r, s, t, 5 are in GP, find the value of p, q, r, s and t.
1
v) If – 64 , x, y, z and –4 are in GP, find the value of x, y and z.

8. i) If 5 , a, b, c, d, 40 and e are in GP, find the value of a, b, c, d and e.
4
IHf opw, q m, 2an7y, r g, eso, tm, aentrdic 13me aarnes ianr eG tPh, efirned b tehtwe eveanlu 5e aonf dp 1, q2,8 r0, sw ahnedre i tt.he third mean is 40?
ii) 1
iii) 9

iv) How many GM's are there between and 729 where the ratio of third mean and
last mean is 1: 81?
v) Find the number of GM's between the terms 3 and 192 where the ratio of the first
mean and last mean is 1:16?

9. i) The 6th term of a GP is 32 times the first term and 9th term is 1280. Find the sequence.
ii) If x + 1, 12 and 5x – 1 are in GP, find the value of ‘x'. Also find the common ratio.
iii) If product of first 3 terms of a GP is 216, find the Second term.
iv) If third term of a GP is 4, find the product of the first 5 terms.
v) There are 8 baskets having full of eggs in GP in each baskets respectively. If 4th and

6th baskets have 54 and 486 eggs respectively find the eggs in first and last baskets.
10. PRIME more creative questions:
i) If (m + n)th term and (m – n)th terms of a GP are p and q respectively, find the mth

term of the sequence.
ii) If pth, qth and rth term of a GP are x, y and z respectively prove that xa – r. yr – p . zp - q = 1.
iii) If 4th, 7th and 10th terms of a GP are x, y and z respectively, prove that y2 = zx.
iv) The first three terms of a GP are x + 6, x and x – 3. Find the fifth term of GP.
Tmheearne aarree i‘nn 't gheeo rmateiotr 9ic: 4m, efiannds t bheet vwaeluene o12f 67‘n'a. nd 243
v) 16 where (n – 1)th mean and 4th

Answer

1. Show to your teacher. 2. Show to your teacher.
3. i) –16, –64, –256 ii) 768 iii) i1ii6)13 8414
iv) 2048 v) 5
iv) 9th term v) 11
4. i) 10 ii) 8
v) 5
5. i) 21526800 + 1280 i+i) 6 401 1+ 320 + ... + i.i.i.) + ... 2143 , 1 , 1 , 1 , v13) , 1,4 ..9.,1..5.,2...
iv) 81 27 9 v) 5
2,4374
6. i) 20 ii) 36 iii) ar4 9 iv) x2 – y2
1 27 2 3 1 1
7. i) 4 ii) 2 , 1, 2, 4 iii) 2 , , 2 , 2 , 6

iv) 160, 80, 40, 20, 10 v) – 1 , – 1 , –1
5 16 4
2
8. i) , 5, 10, 20, 80 ii) 243, 81, 9, 3, 1 iii) 7 iv) 7

9. i) 5, 10, 20, ..., ..., ... ii) 5, 2 iii) 6 iv) 1024 v)
3
10. i) pq iv) 4 v) 7

58 PRIME Opt. Maths Book - X

Sum of the terms of GP

Let us consider the sequence 2, 6, 18, 54, ..., ..., ... is in GP where,
First term (a) = 2
t2 6
Common ratio (r) = t1 = 2 = 3.

Then, nth term is,

t n = ar n – 1
= 2 × (3) n – 1

Here,

The sum of the terms of GP is, 2 + 6 + 18 + ... + ... + ... which can be written in sigma
/k /k

notation as (Sn) = tn = 2 3n – 1
n=1
n=1

If there are only four terms in the sequence the sum can be calculated by expressing in

expanded form

i.e. Sn /k

= 2 3n – 1

n=1

= 2 × 31 – 1 + 2 × 32 – 1 + 2 × 33 – 1 + 2 × 34 – 1
= 2 + 6 + 18 + 54
= 80

Derivation of formula of Sn

Let us consider the terms of a GP as,
MSnu =l taip +ly a rb y+ ‘ar'r,3 w +e a gre4 t+, ... + ... + ... + arn – 1 .................................................... (i)
Subrtrsan c=t ianrg + e aqr (2i +) aanr3d + ( iai)r,4 w +e .. .g +e t.... + ... arn ................................................ (ii)
srsn n= =( a– a)+r ar + ar2 + ar3 ... + ... + ... + arn –1
+ ar2 + ar3 + ... + ... + ... + arn –
(–) (–) (–) (–) 1 + arn
(–)

or, SSnn (=1 a–1 –r–)a r=rn a – arn

or, Sn = a (rn – 1) Where r !1.
r–1

The calculating formula of sum of the terms in GP is,
a (rn – 1)
Sn = r–1 , r !1

Also, a (rn – 1)
r–1
Sn =

or, Sn = arn –1 r – a
r–1

or, Sn = tn . r – a
r–1

= rtn – a Where tn is the last term.
r –1

PRIME Opt. Maths Book - X 59

Worked out Examples

1. Find the sum of the first 9 terms of the sequence 2, 4, 8, + ... + ... + ... + ... .

Solution:

First term (a) = 2 t2 4
t1 2
Common ratio (r) = = = 2

Sum of first 9 terms (s9) = ?
We have,
a (rn – 1)
Sn = r–1

or, S9 = 2 (29 – 1)
2–1

or, S9 = 2 × 511
1

\ S9 = 1022

2. Find the sum of the terms of the series 1 + 1 + 1 + ... + ... + ... + ... + 27.
9 3
Solution:
Alternative method:
First term (a) = 1 or, 2tn7 = = a r19 n – × 1 3n – 1
9 or, 35 = 3n – 1
\ n = 6
1

Common ratio (r) t2 = 3 = 3.
last term (tn) = 27. t1
1
We have, 9

Sn = tn . r–a Again,
r –1

= 27 × 3 – 1 Sn = a^rn – 1h
3–1 9 r–1

= 729 – 1 S6 = 1 ^36 – 1h
2×9 9
\
3 –1

= 364 = 1 × 728
9 9 2

= 40 4 . = 364
9 9

= 40 4 .
9

3. If sum of first 2 terms of a GP is 9 and sum of first 4 terms is 45, find the first term and

common ratio. Also find the 6th term.

Solution:
SS42 == 945
We have,

Sn = a (r2 – 1)
r–1

or, S2 = a (r2 – 1)
r–1

60 PRIME Opt. Maths Book - X

or, 9 = a (r2 – 1) ....................................... (i)
r–1

Again,

S4 = a (r4 – 1)
r–1

or, 45 = a (r2 + 1) (r2 –1) ....................................... (ii)
r–1

Dividing equation (ii) by (i),

or, 45 =
9

or, 5 = r2 + 1
or, r2 = 4
\ r = 2

Substituting the value of 'r' in equation (i),

9 = a (22 – 1)
2–1

or, 9 = 3a
\ a = 3

Again,
tt n6 === 33ar ××n –22 156 – 1
= 3 × 32
= 96

\ First term (a) = 3
Common ratio (r) = 2

6th term (t6) = 96

PRIME Opt. Maths Book - X 61

Exercise 1.3 'D'

1. Answer the following questions.
i) Write down the formula to find sum of first 'n' terms of a geometric sequence.
ii) Write down the formula to find sum of first 'n' terms of a GP having last term.
iii) Are the terms a , a and ar in GP?
r
iv) Are the terms a a, ar, ar3 in GP?
r3 , r

v) What will be the 'nth' term of the sequence a, ar, ar2, ..., ...,... ?

2. Write down the followings in expanded form and find the sum of the terms.
/4 /5 /5

i) 3 2n –1 ii) 2 4n –1 iii) 5 (–2)n +1
n=1 n=2 n=1

iv) Find the nth term of the series 5 + 10 + 20 + 40 + 80. Also write down in sigma
notation.

v) Write down the given series in sigma notation by calculating the nth term of 2 – 6 +
18 – 54 + 162.

3. Find the sum of the following:
i) 1 + 2 + 4 + 8 + ... + ... + ... + 9 terms.
ii) 81 + 54 + 36 + ... + ... + ... + 8 terms.
1
iii) 2 + 1 + 2 + ... + ... + ... +10 terms.

iv) 3 + 6 + 12 + ... + ... + ... + 1536.
1 2
v) 9 + 3 + 4 + ... + ... + ... + 144.

4. i) The 3rd term and 7th term of a GP are 16 and 256 respectively. Find the sum of the
first 10 terms.

ii) The 4th term and 7th term of a GP are 80 and 10 respectively. Find the sum of the
first 8 terms.

iii) If 5th term of a GP is 8 times the second term and 6th term is 96, find the sum of the
first 10 terms.

iv) The sum of the first 2 terms and first 4 terms of a GP having positive common ratio
are 6 and 30 respectively. Find the sum of the first 8 terms.

v) The sum of the first 3 terms and first 6 terms of a GP are 1 and 28 respectively. Find
the sum of the first 10 terms of the sequence.

5. i) The sum of the first two terms of a GP is 18 and sum of the next two terms is 162.
Find the first term and positive common ratio.

ii) The sum of the first 6 terms of a GP is 9 times the sum of the first three terms. Find
the common ratio.

iii) If 3rd term and 6th term of a GP are in the ratio 1 : 8 where the 8th term is 384. Find
the sum of the first 8 terms.

iv) If k + 9, k – 6 and 4 are in GP find the sum of the first 5 terms.
v) The sum of the first 4 terms and last 4 terms of a GP having 8 terms are 45 and 720

62 PRIME Opt. Maths Book - X

respectively. Find the first term and common ratio.

6. i) If nth term of a GP 5, 10, 20, ..., ..., ..., and nth term of another GP 1280, 640, 320 ..., ...,
..., are equal. Find the number of terms.

ii) Find the three terms in GP whose sum is 21 and the product of their extremes is 36.
iii) Find the three terms in GP whose product is 1000 and sum of their squares is 525.
iv) The first term of a GP is 2. If sum of 3rd term and 5th term is 40, find the common

ratio.
v) If sum of the terms of series. 2 – 2 2 + 4 – 4 2 + ... + ... + ... + tn is 2(15 – 7 2 ), find

the value of tn.

7. PRIME more creative questions.
i) Pranav borrows Rs. 4368 where he promises to pay in 6 equal installments, each

installment being triable of the preceding installment. Find the first and last
installment.
ii) Pranisha Rs. 400 in one month, Rs. 800 in second month and Rs. 1600 in 3rd month
in a bank. Find here saving amount int he bank after 8 months.
iii) Find the sum of the first 5 terms of a GP in which the third term is greater than the
first by 9 and the second term is greater than the fourth by 18.
iv) Find the product of the five consecutive even terms of GP where the 6th term of GP
is 4.
v) The sum of the first 4 terms of a GP is 30 and sum of last 4 terms is 960 having first
term 2 and last term 512. Find the number of terms .

Answer

1. Show to your teacher. ii) 680 iii) 220

2. i) 45 /5

/5 v) 2 (–3)n+1

iv) 5 2n–1 n=1

n=1

3. i) 511 ii) 6305 iii) 1023 iv) 3069 v) 1555
27 256 9

4. i) 4092 ii) 1275 iii) 3069 iv) 510 v) 29524
13

5. i) 9 , 3 ii) 2 iii) 765 iv) 1031 v) 3, 2
2 ii) 3, 6, 12 or 12, 6, 3 25
iv) 2
6. i) 5 ii) Rs. 1,02,000 v) 16
iii) 5, 10, 20 or 20, 10, 5 v) 9 iii) 93

7. i) Rs. 12, Rs. 2916
iv) 1024

PRIME Opt. Maths Book - X 63

Sum of natural numbers and relationship between AM and GM.

1. Sum of first ‘n' natural numbers:
The first ‘n' natural numbers are taken as 1, 2, 3, 4, ..., ..., ..., upto n terms.

Here,
First term (a) = 1
Common difference (d) = t2 – t1 = 2 – 1 = 1.

We have,

Sum of the terms,
n
Sn = 2 [2a + (n – 1)d]

= n [2 × 1 + (n – 1)1]
2

= n [n + 1]
2

\ Sn = n (n + 1)
2

2. Sum of the first ‘n' odd number:
The first ‘n' odd number are 1, 3, 5, 7, ..., ..., ..., n terms

Here,
First term (a) = 1
Common difference (d) = t2 – t1 = 3 – 2 = 2

We have,

The sum of the terms is
n
Sn = 2 [2a + (n – 1) d]

= n [2 × 1 + (n – 1) 2]
2

\ Sn = n [2 + 2n – 2]
2

= n × 2n
2

\ Sn = n2

3. Sum of the first ‘n' even numbers.
The first ‘n' even numbers are taken as 2, 4, 6, 8, 10, ..., ..., ..., upto n terms, written in

series.

Here,
First term (a) = 2
Common difference (d) = t2 – t1 = 4 – 2 = 2.
Then, sum of the terms is ,
n
Sn = 2 [2a + (n – 1) d]

= n [2 × 2 + (n – 1) 2]
2

= n [4 + 2n – 2]
2

\ Sn = n(n + 1)

64 PRIME Opt. Maths Book - X

4. Sum of the square of first ‘n' natural numbers.
Let us consider the sum of the square of first ‘n' natural numbers are,
WSn e= k 1n2o +w 2 2 +n 33 –2 +(n .. .– + 1 .)..3 + = . ..n +3 n–2 n3 + 3n2 – 3n + 1 = 3n2 – 3n + 1
Putting n = 1, 2, 3 ..........................n,

We have
13 – (1 – 1)3 = 3 × 12 – 3 × 1 + 1
23 – (2 – 1)3 = 3 × 22 – 3 × 2 + 1
33 – (3 – 1)3 = 3 × 33 – 3 × 3 + 1
(n – 1)3 – (n – 2)3 = 3 × (n – 1)2 – 3 × (n – 1) + 1

n3 – (n – 1)3 = 3n2 – 3n + 1

Adding all of the above, we have
n3 – 03 = 3(12 + 22 + 33 + ... + ... + ... + n2) – 3(1 + 2 + 3 + ... + ... + ... +n) + (1 + 1 + 1 + ...+ ... + ... + 1)

n3 = 3.Sn – 3n (n + 1) + n n terms
2

or, n3 + 3n (n + 1) – n = 3Sn
2

or, (n3 – n) + 3n (n + 1) = 3Sn
2

or, n(n + 1) (n – 1) + 3n (n + 1) = 3Sn
2

or, n(n + 1) [n – 1 + 3 ] = 3Sn
2

or, = 3Sn

\ Sn =

Relation between AM and GM

Let AM and GM be the arithmetic and geometric mean between any two positive numbers
‘a' and ‘b'.
a +b
Here, AM = 2
GM = ab

Now,

AM – GM = a + b – ab
2

or, AM – GM = a + b –2 ab
2

or, AM – GM = 1 [ ^ a h2 + ^ bh2 – 2 a. b]
2

or, AM – GM = 1 ^ a– b h2
2

or, AM – GM > 0 (i.e. always positive as square of numbers)
\ AM > GM

PRIME Opt. Maths Book - X 65

Worked out Examples

1. Find the sum of the first 20 even numbers.

Solution:
No. of terms (n) = 20

For even numbers,

\ SS n20 === n2200(n( ×2 + 02 1 1+) 1)

= 420

2. Find the sum of the square of first 15 natural numbers.

Solutions:
No of terms (n) = 15
Sum of the square of first ‘n' natural numbers,

Sn =

=

= 15 5 × 16 8 × 31
6
2

= 1240.

3. Find the 3 terms in AP whose sum is 15. If 1, 4, and 19 are added to them, the terms will be
in GP.
Solution:

Let, the 3 terms in AP be, a – d, a and a + d.
1st condition,
a – d + a + a + d = 15
or, 3a = 15

` a = 5

2nd Condition:
The terms a – d + 1, a + 4 and a + d + 19 are in GP
or, 6 – d, 9 and 24 + d are in GP,
t2 t3
t1 = t2

or, 6 9 = 24 + d
–d 9

or, 81 = 144 + 6d – 24d – d2
or, d2 + 18d – 63 = 0
or, d2 + (21 – 3)d – 63 = 0
or, d2 + 21d – 3d – 63 = 0

or, d(d + 21) – 3(d + 21) = 0
or, (d + 21) (d – 3) = 0

Either OR
d + 21 = 0 d – 3 = 0
\ d = – 21 \ d = 3

66 PRIME Opt. Maths Book - X

Then,
a – d = 5 – 3 or 5 –(– 21) = 2 or 26
a = 5
a + d = 5 + 3 or 5 + (–21) = 8 or – 16
\ The required terms are 2, 5, 8 or 26, 5, – 16.

4. Find any two numbers whose AM and GM are 25 and 24 respectively.

Solution:
Let, ‘a' and ‘b' be any two numbers
AM = 25
GM = 24
Now, taking,
a + b
AM = 2

or, 25 = a + b
2

or, a + b = 50

` a = 50 – b ........................................ (i)

Again,
GM = ab
or, (24)2 = ^ abh2
or, 576 = (50 – b) b

or, b2 – 50b + 576 = 0
or, b2 – (32 + 18) + 576 = 0
or, b2 – 32b – 18b + 576 = 0
or, b(b – 32) – 18(b – 32) = 0
or, (b – 32) (b – 18) = 0

Either OR
b – 32 = 0 b – 18 = 0
\ b = 32 b = 18

Substituting the value of ‘b' in equation (i)
a = 50 – 32 = 18
a = 50 – 18 = 32
\ The required terms are 32 and 18 or 18 and 32.

PRIME Opt. Maths Book - X 67

Exercise 1.3 'E'

1. Find the sum of the followings:
i) Sum of first ‘n' natural numbers.
ii) Sum of first ‘n' odd numbers.
iii) Sum of first ‘n' even numbers.
iv) Sum of first 50 natural numbers.
v) Sum of first 20 odd numbers.

2. Find the sum of following:
i) Sum of first 40 even numbers.
ii) Sum of the square of first ‘n' natural numbers.
iii) Sum of the square of first 20 natural numbers.
iv) Find the sum of the terms of 12 + 22 + 32 + 42 + ... + ... + ... + 102
v) Sum of the terms of square from 1 to 100.

3. i) Find any three terms in AP whose sum is 12. If 1, 2 and 6 are added to them
respectively the terms will be in GP.

ii) Find any three terms in AP whose sum is 15. If 2, 5 and 13 are added to them
respectively, the terms will be in GP.

iii) Find the three terms in GP whose product is 64. If 1, 4 and 5 are added to them, the
terms will be in AP.

iv) If 2nd, 4th and 9th terms of an AP are in GP, find the common ratio of the geometric sequence.
v) Three numbers are in the ratio 1 : 4 : 12 : If one is added to the first number, the

resulting numbers will be in GP, find the numbers.

4. i) Find any two terms whose AM is 50 and GM is 40.
ii) Find any two terms whose AM is 34 and GM is 30.
iii) The Sum and difference of AM and GM of any two numbers are 80 and 20 respectively.

Find the numbers.
iv) If a, b, and c are in AP where x is GM between a & b, y is GM between b & c. Prove that

b2 is the AM between x2 & y2.
v) If ratio of AM and GM between any two numbers is 5 : 3, find the ratio of such

numbers.

5. PRIME more creative questions:
a. i) The sum of any three terms of an AP is 15 where geometric mean of first and third

term is 4. Find the terms.

68 PRIME Opt. Maths Book - X

ii) If a, b and c are in AP and p, q & r are in GP, prove that pb – c qc – a ra – b = 1.
iii) The sum of the first four terms of a GP is 15 and arithmetic mean of first and last

term is 4.5, find the common ratio of the GP.
iv) If product of any two numbers is 160 when arithmetic mean of them is 14, find the

numbers.
v) The first term of an arithmetic and a geometric sequence are equal to 1 and their

second terms are also equal. If 14th term of the AP is 3 times the 3rd term of GP, find
the common difference of the AP.

b. i) Add two more patterns and find the nth term of the sequence so formed of the
number of dots. Also write down in sigma notation from the given patterns.

ii) Add two more patterns and find the nth term of the sequence so formed of the
number of dots. Also find the sum of the dots of first 6 diagrams.

iii) Add two more patterns and find the nth term of the sequence so formed of the
number of dots. Also find the sum of the dots of first 10 diagrams.

iv) The 2nd, 6th and 12th terms of an AP are respectively taken in GP. Find the common
ratio of the geometric progression.

v) Find the 13th term of a GP where product of 10th, 12th, 14th and 16th terms is 1296.

6. Project work
The height of a plant is increased everyday by 2 cm where the height on Sunday is 16cm.

What will be the height of the plant on Friday of the same week? After how many days its
height will be 60 cm?

PRIME Opt. Maths Book - X 69

Answer

1. i) n (n + 1) ii) n2 iii) n(n + 1)
2 v) 400
iii) 2870
iv) 1275 iii) 2, 4, 8 or 8, 4, 2
338350
2. i) 1640 ii) 3, 5, 7 or 18, 5, –8 iii) 10, 90 or 90, 10
iv) 385 v) 3, 12, 36 v) 11
3. i) 2, 4, 6 or 11, 4, –3 ii) 18, 50 or 50, 18
1:9 or 9:1 /5
iv) 5 v) 8, 20 or 20, 8
2 ii) tn = 5n, (5n)
n=1
4. i) 20, 80, or 80, 20 ii) 6
iv) v)
5.a. i) 2, 5, 8 or 8, 5, 2 iv)
n (n + 1) /5 n (n +
b. i) tn = 2 , 2 1)

n=1

/5

iii) tn = (3n + 2), (3n + 2) , 185
n=1

iv) 3 v)

Sequence and series

Unit Test

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20] Time : 30 minutes

Attempt all the questions:

1. Prove that AM between any two terms is a + b .
2

2. a) Is 192 a term of the sequence 3, 6, 12, 24, ..., ..., ..., ?
b) If 12, x, y, z and 28 are in AP, find th value of ‘x', ‘y' and ‘z'.
1 1
c) Find the sum of the terms of the series : 9 + 3 + 1 + ... + ... + ... + 81.

3. a) Find the three terms in AP whose sum is 18. If 1, 3 and 17 are added to them
respectively, the terms will be in GP.

b) How many GMs are there between 1280 and 5 where the first mean and last mean
are in the ratio 64:1?

4. The 5th term of an AP is 52 and sum of first 8 terms is 444, find the series. Also find which
term of the series is 10?

70 PRIME Opt. Maths Book - X

1.4 Introduction of linear programming

Enjoy the recall ( for p & q are equal )
Inequalities:
Taking any two numbers p and q.

Where,

p – q = 0

p – q ≠ 0 (for p & q are unequal)

p – q > 0 ( +ve) (for p > q)

p – q <0 (–ve) ( for p < q)

p ≥ q [ for p = q or p > q]

p ≤ q [ for p = q or p < q]

Here, the signs >, <, ≥ and ≤ are used for inequalities or inequations.

Linear inequality in two variables.
The inequality x > 3 in number line is,

......... -4 -3 -2 -1 0 1 2 3 4 5 .........
The inequality x ≤ 1 in number line is,

......... -4 -3 -2 -1 0 1 2 3 4 5 .........

• In both cases value of the variable x is not fixed, the number line shows the direction
where there are the values of ‘x'.

• The inequality of two variables x and y like ax + by + c < 0, ax + by + c > 0, ax + by
+ c ≤ 0 and ax + by + c ≥ 0 shows the system of linear inequalities of two variables
x and y. Here, we discuss the solution of such in equations.

NOTE:
• For the symbols of inequality ≥ or ≤ (greater or equal or less or equal)
The straight line which represents the boundary line of given
inequality is shown by bold(solid) line like
• For the symbols of inequality > or < (greater or less only)
The straight line which represents the boundary line of given
inequality is shown by liquid (broken or dotted) line like

Note :
i) For any inequality ax + by + c ≥ 0 (or ≤ or < or >), ax + by + C = 0 is the boundary line.
ii) ax + by + c = 0 divides the plane in two equal halves.
iii) Any point satisfying the inequality indicates the plane for solution set.

PRIME Opt. Maths Book - X 71

Ways of obtaining solution set of the inequality Y
i. Solution set of the inequality x ≥ 0.
TP(1, 2)
Here,
x = 0 ( y - axis ) represents the straight line as boundary O X
(x=0)
line. X'
Taking any of the point like (1, 2) as the testing point, the Y'

inequality x > 0 shows 1 > 0 (which is true)

So,
the solution set lies towards the point (1, 2).

ii. x ≤ 2 shows the solution set as: Y (x=2)
O TP(3, 2)
Here, Y' X
x = 2 (parallel to y - axis ) represents a straight line as
X
its boundary line. X'
Taking any of the point (3, 2) as the testing point the
inequality x > 2 gives 3 > 2 (which is false)

So,

the solution set lies in the opposite side of the testing
point(3, 2).

Y

iii. y ≥ 0 shows he solution set as:

Here,
y = 0 (x - axis) represents a straight line as its
boundary line. X'
(y=0) O
Taking (–2, –3) as the testing point, the inequality
becomes –3 > 0 (which is false).
So, TP(–2, –3)

the solution set lies opposite to the point. Y'

iv. Solution set shown by the in equality y ≤ – 3. TP(-1, 1) Y
O
Here,
y = – 3 is the straight line parallel to x - axis presents X' Y'
X
the boundary line. (y=–3)
Taking a point (–1 , 1) as the testing point for y < – 3.
The inequality becomes 1 < – 3.
(It is false)

So,

it's solution set is opposite from the testing point.

Note : Always, we choose a testing point through which the boundary line doesn't pass.

72 PRIME Opt. Maths Book - X

v. Solution set shown by the inequality 2x + y ≤ 0. Y

Here, TP (2, 3)

2x + y = 0 is the straight line represented by the given (-1, 2) X
(1, -2)
inequality as its boundary line.
Where y = – 2x
X' (0, 0) O
x 1 – 1 0

y – 2 2 0

Taking the point (2, 3) as the testing point, the (y= 0) (y= –2x)
inequality becomes 7 < 0 which is false. Y'

So, the solution set is opposite from the testing point.

vi. Solution set shown by 2x + 3y ≥ 12.
Here, The equation 2x + 3y = 12 shows

the straight line which represents the

boundary line given inequality.

y = 12 – 2x Y
3

x036 (0, 4)
y420
(3, 2)

Taking (1, 1) point as the testing point TP(1, 1) (6, 0)
where the inequation becomes, 5 < 12 X' O 2Xx + 3y = 12
(which is false).
Y'
So, the solution set is opposite of the
testing point.

vii. Solution set shown by 2x + 3y ≤ 12.

Here Y
The equation 2x + 3y = 12 shows the

straight line which represents the

boundary line given inequality where (0, 4)

y = 12 – 2x
3
(3, 2)
x063
TP(1, 1)
y402 X' O (6, 0)
2xX + 3y = 12
Taking point (1, 1) as the testing point Y'
where the inequality becomes 5 < 12 (It
is true).

So, the solution set lies towards the

testing point.

PRIME Opt. Maths Book - X 73

Solution set shown by two or more linear inequalities of two variables.
Here, we discuss the solution set of a system of linear inequalities in two variables which is the

intersecting region on a plane represented by the given system.

i. Solution set of the inequalities x + y = ≤ 6 and x – 2y ≥ 0.
Here, x + y = 6 and x – 2y = 0 are the equation of straight lines represented by the given

inequalities as their boundary lines.

The points for the boundary lines are y = x
y = 6 – x 2

x062 x024

y604 y 0 1 2

Taking (1, 1) as the testing point for both where the inequalities for (1, 1) becomes 2 < 6
(It is true) for first inequation –1 > 0 (It is false) for second equation x –2y > 0.

x + y = 6 Y

(0, 6)

(2, 4)

TP(1, 1) (4, 2)

X' (2, 1) (6, 0) X
x – 2y = 0 O (0, 0)

Y'

The solution set lies towards the testing point for first and opposite of testing point for
the second. The intersected solution set is shown in diagram.

ii. The solution set represented by x – y ≥ – 6, 2x + y ≥ 0 and x ≤ 2.
Solution :
Here, the system of inequalities is given by

x – y ≥ – 6 ........................... (i)
2x + y ≥ 0 ........................... (ii)

x ≤ 2 ........................... (iii)
The boundary lines for x – y = – 6 is x – y = – 6 and the points on it are

x 0 – 6 1
y607

74 PRIME Opt. Maths Book - X

Testing at (1, 1) Y B y = x + 6
1 – 1 ≥ –6
i.e. 0 ≥ – 6 (which is true) (1, 7)
\ The half plane of (i) contains (1, 1) (0, 6)

Again, the boundary lien of 2x + y ≥ 0 is 2x + y = 0 C x = 2

and the points on it are

x012 TP(1, 1)
(0, 0) O
y 0 – 2 – 4 X' (–6, 0) X
(1, –2)
Taking at (1, 1) A (2, –4)
2 × 1 + 1 ≥ 0
i.e. 3 ≥ 0 (true)
\ The half plane of (ii) contains (1, 1)

The boundary line of x 2 is x = 2 and the points on Y' y = – 2x
it are
x222
y012

Testing at (1, 1), 1 ≤ 2 (true)
\ The half plane of (ii) contains (1, 1)

The solution set is towards the testing point for all the inequations.

So, the solution set is the intersected region as shown in the diagram from which vertices
of the intersected region are A, B and C.

iii. Solution set represented by the inequations x + y ≤ 6, 3x – 2y ≤ 6, x ≥ 0, y ≥ 0.
Here, the equations x + y = 6, 3x – 2y = 6, x =
0 and y = 0 represents the boundary line for
Y

the given inequalities. y = 6 – x

Where,
y = 6 – x
3x – 6 (0, 6) C
y = 2

x 064 B (4, 2)
y 602 (6, 0)

x 02 4 TP(1, 1) AX
y – 3 0 3 X' (0, 0) O
x = 0
y = 0 (It is y - axis)
(It is x - axis)

Taking (1, 1) as the testing point for all Y'

PRIME Opt. Maths Book - X 75

We have
2 < 6 (It is true)

1 < 6 (It is true)

1 > 0 (It is true)

1 > 0 (It is true)

So, the solution set lies towards the testing point for all the inequalities where the
vertices of intersected set becomes O, A, B & C.

1.4.1 Linear programming

The maximum and minimum values of the function F = ax + by + c is found out by taking
the vertices of the intersected solution set of the given inequalities which is also called
optimization of the function. This can be done in different fields using optimization in the
topic of linear programming.

It can be used to find out the optimum (maximum & minimum) value in different fields like
to find the maximum cost and minimum cost of the goods in different market after sampling
the data, to find the maximum and minimum salary of the civil servants in different private
companies, maximum and minimum size of shoes produced by factories according to the
collection of data among the consumers, etc for maximum profit.

The process of finding out the maximum and minimum (optimum)
values of a function F = ax + by + c by using the vertices of the
solution set given by the inequalities is called linear programming.

Terms used in linear programming
i. Objective Function:
The function which is used to find out the maximum or minimum values like p = ax + by

where x and y represents the unit cost of different units as an example is called objective
function.

ii. Constraints:
The inequalities which indicate the different units taken in an observation to find the

values of the variables of a function are called constraints.

iii. Feasible region:
The area bounded by the straight lines represented by the given inequalities is called
feasible region. The variables x and y are taken from the vertices of feasible region to find
the value of the objective function.

76 PRIME Opt. Maths Book - X

Example
Maximize the function F = 3x + 2y + 3 under the constraints x + y ≤ 6, 2x – y ≤ 6, x ≥ 0 and

y ≥ 0.
Solution:

The corresponding straight lines for the given constants are
x + y = 6 ......................................... (i)
2x – y = 6 ....................................... (ii)
x = 0 ................................................. (iii)
y = 0 ................................................. (iv)

From equation (i) From equation (ii)
y = 6 – x y = 2x – 6

x061 x031
y605 y – 6 0 – 4

From equations (iii) and (iv)
x = 0 ( It is y - axis ) y = 0 ( It is x - axis )

Let, (1, 1) be a point taken as the testing point for all inequalities.

Inequalities Value at (1, 1) Result Remarks

x + y < 6 2 < 6 True Towards TP

2x – y < 6 1 < 6 True Towards TP

x > 0 1 > 0 True Towards TP

y > 0 1 > 0 True Towards TP

Y

x = 0
2x – y = 6
C(0, 6) (1, 5)

TP(1, 1) B(4, 2)
(6, 0)
X' (0, 0) A(3, 0) X
O y = 0

x + y = 6

(1,-4)

Y' 77
PRIME Opt. Maths Book - X

From graph, we get,
The vertices of the intersected solution Set (feasible region) are 0(0, 0), A(3, 0), B(4, 2)

and C(0, 6).
Then, Taking the function,
F = 3x + 2y + 3

At

Vertex Objective function F = 3x + 2y + 3 Remarks

O(0, 0), F = 3 x 0 + 2 x 0 + 3 = 3 minimum

A(3, 0), F = 3 x 3 + 2 x 0 + 3 = 12

B(4, 2), F = 3 x 4 + 2 x 2 + 3 = 19 maximum

C(0, 6), F = 3 x 0 + 2 x 6 + 3 = 15

Here, The maximum value of the function ‘F' at point (4, 2) is 19.

Worked out Examples

1. Minimize the function P = 5x + 2y under the constraints 3x + 4y ≤ 24, x ≥ – 2 and x – 2y ≤ 0.
Solution:
The equation of straight lines corresponding to the given inequalities are:

3x + 4y = 24 .................................(i)
x – 2y = 0 ......................................(ii)
x = – 2 ............................................(iii)

From equation (i) From (ii)
24 – 3x x
y = 4 y = 2
4
x08 3 x 0 2 –2
0 1 –1
y60 y

From equation (iii)
x = – 2 (It is parallel to y-axis )

Let, the point (0, 2) is taken as the testing point for all the inequalities.

Inequations Value at (0, 2) Result Remarks
3x + 4y < 24 8 < 24 True Towards TP
x – 2y < 0 – 4 < 0 True Towards TP
x > – 2 0 > – 2 True Towards TP

From graph, we get,
The vertices of feasible region of the inequalities are A(– 2, – 1), B(4, 4.8) by solving

equation i and ii and C(– 2, 7.5) by solving equation ii and iii.

78 PRIME Opt. Maths Book - X

C(–2, 37x. 5+ )4y = 24 Y
(0, 6)

x = – 2 (4, 3)

B(4, 4.8)

TP(1, 1) (2, 1)

X' (–2,–1) O (0, 0) (8, 0) X
x – 2y = 0 A
Y'

Taking the function p = 5x + 2y,
At A(– 2, – 1), p = 5(– 2) + 2( –1) = – 12 (minimum)
B(4, 2), p = 5 x 4 + 2 x 2 = 24
C(– 2, 7.5) p = 5 x (– 2) + 2 x 7. 5 = 5
\ – 12 is the minimum value of the function at point (– 2, – 1)

Note : If graphs donot intersect at the point where abscissa and ordinate are both integer, better
to solve the equations to get vertices.

2. Optimize the function Z = 2x + 4y under the constraints x – y ≥ – 6, x + y ≥ – 2, x ≤ 0, y ≤ 0.
Solution:
The equation of straight lines corresponding to the inequations are:

x – y = – 6 .....................................(i)
x + y = – 2 .....................................(ii)
x = 0 ...............................................(iii)
y = 0 ................................................(iv)

From equation (i) From equation (ii)
y = x + 6 y = – x – 2
–2
x 0 – 6 4 x0 2 –2
y60 y – 2 – 4 0

From equation (iii) and (iv)
x = 0 (It is y - axis)
y = 0 (It is x - axis )

PRIME Opt. Maths Book - X 79

Let, (1, 1) be a testing point for all the inequalities where,

Inequalities Value at (1, 1) Result Remarks

x – y > 6 0 > – 6 True Towards TP

x + y > –2 2 > – 2 True Towards TP

x < 0 1 < 0 False Opposite to TP

x < 0 1 < 0 False Opposite to TP

Y x – y = – 6
C (0, 6)

(–2, 4) x = 0

B(–4, 2) TP(1, 1)
(0, 6)
X' A O (0, 0) y = 0 X

(0, –2)

(2, –4)

Y' x + y = – 2

From graph, we get,
The vertices of feasible region are O(0, 0), A(– 2, 0), B(–4, 2) and C(0, 6).

Then, For the function Z = 2x + 4y
At point O(0, 0), Z = 2 x 0 + 4 x 0 = 0
A(–2, 0), Z = 2(–2) + 4 x 0 = –4 (Minimum)
B(–4, 2), Z = 2(–4) + 4 x 2 = 0
C(0, 6), Z = 2 x 0 + 4 x 6 = 24 (Maximum)
\ –4 is minimum value at point A(–2, 0) and 24 is maximum at point C(0, 6).

80 PRIME Opt. Maths Book - X

3. From the given feasible region find the inequalities where one of the inequality is x + y ≤ 6
and two vertices are B(–2, –1) and C(–2, 8). Also maximize the function P = 2x + 3y – 1.
Y

C(–2, 8)

A(4, 2) X

X' O
B(–2, 1)

Y'
Solution:
Here, the feasible region is bounded by the sides AB, AC and BC where the given inequality
is, x + y ≤ 6.

It represents the straight line AC which cuts x-axis at (6, 0) and y-axis at (0, 6). Two
vertices are B(–2, 8) and C(–2, –1).

Using two points formula,

y – y1 = y2 – y1 (x – x1)
x2 – x1

Equation of OB Equation of BC is,
–1 –0 8+1
y – 0 = –2 –0 (x – 0) y + 1 = –2 + 2

or, 2y = x or, x + 2 = 0
or, x – 2y = 0 or, x = –2

Solving equation of OB : x = 2y and AB : x + y = 6
x = 4 and y = 2
\ Co - ordinate vertex A is (4, 2).
Let, (0, 2) be a testing point to find the inequalities.

Equations Value at (0, 2) Inequalities

x + y = 6 2 < 6 x + y ≤ 6

x – 2y = 0 – 4 < 0 x – 2y ≤ 0

x = –2 0 > – 2 x ≥ – 2

PRIME Opt. Maths Book - X 81

Again, For the function P = 2x + 3y –1,
At A(4, 2), P = 2 x 4 + 3 x 2 – 1 = 13
B(–2, –1), P = 2(–2) + 3 (–1) – 1 = –8
C(–2, 8), P = 2(–2) + 3 x 8 – 1 = 19 (maximum)
\ 19 is maximum value at point C(–2, 8).

4. Find the inequalities of the given feasible region where the vertices are A(2, 0), B(4, 3) and
C(0, 5). Also minimize the function P = 4x + 3y.
Y

C(0, 5)

B(4, 3)

X' O A(2, 0) X

Y'
Solution:
The given feasible region is bounded by the sides AB, BC, OA and OC where the given
vertices are A(2, 0), B(4, 3) and C(0, 5).

Using two points formula,
y2 – y1
y – y1 = x2 – x1 (x – x1)

For line AB, For line BC,
3–0 3–5
y – 0 = 4–2 (x – 2) y – 5 = 4–0 (x – 0)

or, 2y = 3x – 6 or, 2(y – 5) = – x
or, 3x – 2y = 6 or, x + 2y = 10

For line OA, (It is x - axis) For line OC, (It is y-axis)
y = 0 x = 0

Let, (1, 1) be a testing point to find inequations.

Equations Value at (1, 1) Inequations

3x – 2y = 6 1 < 6 3x – ≤ 6

x + 2y = 10 3 < 10 x + 2y ≤ 10

y = 0 1 > 0 y ≥ 0

x = 0 1 > 0 x ≥ 0

82 PRIME Opt. Maths Book - X

Again, For the objective function p = 4x + 3y.
At point O(0, 0), P = 4 x 0 + 3 x 0 = 0 (minimum)
A(2, 0), P = 4 x 2 + 3 x 0 = 8
B(4, 3), P = 4 x 2 + 3 x 3 = 17
C(0, 5), P = 4 x 0 + 3 x 5 = 15

Here,
O is the minimum value at point O(0, 0).

Exercise 1.4

1. Answer the following questions.
i) Write down the inequality represented by the line x + 2y = 12 for the point (1, 2).
ii) What do you mean by feasible region?
iii) What do you mean by objective function?
iv) Find the equation of straight line joining the points (1, 2) and (3, 8).
v) Is the point (1, 1) satisfied the inequality 2x + 3y = 6?

2. Manimize the given objective functions under the given constraints.
i) P = 5x + 3y, x + y ≤ 6, x – y ≤ 4, x ≥ 0 and y ≥ 0.
ii) P = 4x + 2y, 2x + 3y ≥ 6, 2x – 3y ≤ 6 and y ≤ 2.
iii) F = 3x + 5y, x + y ≥ 10, 2x + 3y ≤ 24, x ≥ 0 and y ≥ 0.
iv) Z = 4x + 3y, 2x – y ≥ 0, x + 2y ≥ 5 and x ≤ 5.
v) C = 3x + 4y, x + 2y ≤ 6, 3x + 2y ≤ 12, x ≥ 0 and y ≥ 0.

3. Maximize the given objective functions under the given constraints.
i) Z = 3x – 2y, x + y ≤ 3, x ≤ 2, y ≤ 1, x ≥ 0 and y ≥ 0.
ii) P = 5x – y, x + y ≤ 0, x – y ≤ 0, x ≥ –3 and y ≤ 0.
iii) F = 3x + 2y + 5, 3x – y ≥ 3, x + y ≤ 5, x ≥ 0 and y ≥ 0.
iv) P = 7x + 3y, x + y ≤ 7, x + 2y ≥ 10, x ≥ 0 and y ≥ 0.
v) C = 9x + 5y, x – 2y ≤ 7, x + 2y ≥ –1, and y ≤ 0.

4. Optimize the given objective functions under the given constraints.
i) P = 2x + 3y, 3x + 2y ≥ 12, 7x + 2y ≤ 28 and x ≥ 0.
ii) F = 2x + 5y, x – 2y ≤ 1, x + y ≤ 4 and x ≥ – 3
iii) P = 7x + 6y, x + y ≤ 6, x – 5y ≤ 0 and 5x – y ≥ 0.
iv) Z = 3x + 2y, x + y ≤ 8, x + y ≥ 6, x – y ≤ 0 and x ≥ 0.
v) F = 3x + 5y – 2, 6x – 5y ≤18, 3x + 2y ≥ –18, x ≤ 0, y ≤ 0.

PRIME Opt. Maths Book - X 83

5. PRIME more creative questions:
i) Find the inequalities which represents the given feasible region where A(3, 2) and

B(0, 5) are the any two vertices of the feasible region. Also maximize function c = 3x
+ 2y.

Y
B(0, 5)

B(3, 2)

X' O X

Y'

ii) Find the inequalities of the given feasible region where A(3, 0), B(4, 2) and C(0, 6)
are the vertices of the feasible region. Also maximize the function Z = 5x + 3y – 2.

Y
C

B

X' O A X

Y'

iii) Find the inequations from the given feasible region where one of the inequality is
x + y ≤ 6 and two of the vertices are B( – 3, 8) and C(–3, – 2). Also minimize the
function F = 7x + 2y – 3.

Y

B(–3, 8)

A

X' O X
C(–3, –2)

Y'

iv) Find the inequations from the given feasible region where the vertices of feasible
region are A( – 6, 0), B(– 4, 3) and C(0, 5).

Y

C
B

X' A OX

Y'

84 PRIME Opt. Maths Book - X

v) Find the inequalities which represents the given feasible region where the vertices
of feasible region are A(3, 3), B(4, 4), C(0, 8) and D(0, 6).

Y

C

DB

A

X' O X

Y'

Answer

1. Show to your teacher.

2. i) 28 at (5, 1) ii) 28 at (6, 2) iii) 40 at (0, 8)
iv) 50 at (5, 10) 3 iii) 5 at (0, 0)
v) 15 at (3, 2 )
3. i) –2 at (0, 1)
iv) 0 at (0, 0) ii) –15 at (–3, 0)
v) –9 at (–1, 0)

4. i) 8 at (4, 0) and 42 at ii) –16 at (–3, –2) and 29 at (–3, 7)
iii) 9 at (0, 0) and 41 at (5, 1) iv) 12 at (0, 6) and 20 at (4, 4)
v) –38 at (–2, –6) and – 2 at (0, 0)

5. Discuss with your teacher.

PRIME Opt. Maths Book - X 85

1.5 Solve of Quadratic Equation

The second degree, polynomial equation (or function) in
the form of ax2 + bx + c= 0 is called the quadratic equation
in the variable ‘x' where a, b & c are any constants and a≠0.

Enjoy the recall:
• The standard form of the quadratic equation is ax2 + bx + c = 0
• It is the second degree polynomial equation (Algebraic function).
• There are two values of the variable ‘x' are obtained after solving which are called

the roots of the variable i.e., x = –b ! b2 – 4ac
2a

• It can be solved by using different methods which are already discussed in grade IX
in compulsory mathematics like factorization method, square completing method
and using formula method.
• The solve of quadratic equation ax2 + bx + c = 0 to establish the formula is as
follows: ax2 + bx + c = 0
or, ax2 + bx = – c
bx c
or, x2 + a = – a [ \ Dividing both sides by a ]

or, x2 + 2.x. b + a b 2 = a b 2 – C
2a 2a 2a a
k k

or, ax + b 2 = b2 – 4ac
2a 4a2
k

or, ax + b k = ! b2 – 4ac
2a 2a

or, x = – b ± b2 – 4ac
2a 2a

\ x = –b ! b2 – 4ac
2a

Alternative Method

a+unfbz] sf] 9fsf (750 AD) df hGdg' ePsf ul0ft1 >L >Lw/ cfrfo{n] ju{ ;lds/0f (Quadratic

Equation) sf] xn lgDg z]nf]s dfkm{t lbg' ePsf] 5 .

rt/' fxtju;{ dM} ?k}M kIfåo+ u'0fot] \ .

ax2 + bx + c = 0 cWoQmju{ ?k}oQ' mf} kIff} ttf] dn" d\ ..

or, 4a(ax2 + bx + c) = 4a × 0
or, 4a2x2 + 4abx = – 4ac
or, (2ax)2 + 2.2ax.b + b2 = b2 – 4ac
or, (2ax + b)2 = b2 – 4ac
or, 2ax + b = ! b2 – 4ac
or, 2ax = – b ! b2 – 4ac
x = – b !
\ b2 – 4ac
2a

86 PRIME Opt. Maths Book - X

The meaning of the shloka is as below.
Stanza I : Multiply the equation by four times of the co-efficient of square of unknown
(variable) i.e. by 4a multiply the equation.
Stanza II : Add both sides of the equation by the square of co-efficient of unknown (variable)
and take square roots of both sides.
i.e. add both sides b2 and take possible roots to solve as linear equation.

Here, in Grade ‘X' we are going to discuss the solution of equation using graphical method
only.

The graph of quadratic equation y = ax2 + bx + c, a ≠ 0 is curve which is closed at one side and
open at the another end in vertical position in the parabolic shape.

The parabola which is open at upward direction is called positive and down is called negative

parabola.
• The turning point at the closed side is called the vertex of the parabola.

• The parabola cuts the x-axis at two points at A and

B.

• At the vertex point, b2 – 4ac = 0 as A and B Y
coincide each other. y = ax2 + bx + c

Hence x = – b at the vertex. b
2a 2a

–

Where, x =
y = ax2 + bx + c
2 X' BO AX

= a a– b k + b a– b k + c
2a 2a

b2 – 2b2 + 4ac a– b , 4ac – b2 k
4a 2a 4a
=

\ y = 4ac – b2 Y' vertex of parabola
4a (Turning point)

\ Coordinate of vertex is a– b , 4ac – b2 k
2a 4a

• The straight line passes through the vertex of parabola parallel to y-axis is called the
axis of the parabola.
b
• Equation of axis of parabola is x = – 2a which divides the curve of parabola in two
equal halves.

PRIME Opt. Maths Book - X 87

Different form of the graph of quadratic functions.
1. In the form of y = ax2 where a is any constant and a ≠ 0.
In the form y = ax2
or, y = ax2 + 0.x + 0

For the vertex of parabola,

x = – b = – 0 = 0 &
2a 2a

y = ax2 = a(0)2 = 0

Y y = ax2

X' O X
Y'

i) When ‘a' is positive the curve of y = ax2 is,
Y y = ax2

X' B O A X
Y'

ii) When ‘a' is negative, the curve of y = – ax2 is,

Y

X' B O A X

y = –ax2

Y'

iii) As the value of constant ‘a' becomes more than 1, the width of the curve goes on
decreasing as ‘s' increasing.
Y y = 2x2

width = less

X' B O A X
Y'

88 PRIME Opt. Maths Book - X

iv) As the value of constant ‘a' becomes less than ‘1', the width of the curve goes on increasing

as its value decreasing.

Y y = 1 x2
2

width = more

X' B O A X

Y'

2. In the form of y = ax2 + bx + c.
For the function y = ax2 + bx + c
b2
Vertex = a– b 4ac – k
2a , 4a

So, the nature of the curve of parabola are as follows.

YY

X' AB X X' A BO X
O

P P

Y' Y'

Y Y

P

X' O X X' A B X
PB O
A

Y' Y'

The curve of parabola given in above diagrams conclude that the parabola of quadratic
function is intersected by x-axis at the points A and B where the value of x co-ordinate
of the points A and B are called the roots of parabola (quadratic equation) and P is the
vertex of the parabola.

The coordinate of the points A and B are given below.

A is ( –b + b2 – 4ac , 0) and B is ( –b – b2 – 4ac , 0)
2a 2a

PRIME Opt. Maths Book - X 89

3. In the form of algebraic function y = ax3

Where,

i) y = x3 Y ii) y = – x3 Y

y = x3

X' X X' X

Y' y = – x3

iii) y = 2x3 Y'
Y y = 2x3
less iv) y = 1 x3
2
X' X
Y y = 1 x3
Y' 2

[As the constant ‘a' more
increases, the width of
the curve with y-axis X' X
becomes less]
Y'

[As the constant ‘a' decreases, the width
of the curve with y-axis becomes more]

Worked out Examples

1. Draw the graph of function y = x2

Solution:

The given function is,
y = x2
or, y = 1.x2 + 0.x + 0

Here, vertex of the parabola of it

Vertex = a– b , 4ac – b2 k
2a 4a

= (0, 0)
Then, y = x2

The points satisfying the equation are

x –3 –2 – 1 0 1 2 3

y 94 1 0149

90 PRIME Opt. Maths Book - X

2. Draw the graph of y = –2x2

Solution:

The given function is,

y = – 2x2
or, y = – 2x2 + 0.x + 0

Here,

Vertex = a– b 4ac – b2 k
2a , 4a

= (0, 0)

Then,
y = – 2x2

The points satisfying the equation are
x – 3 – 2 –1 0 1 2 3
y – 18 – 8 –2 0 – 2 –8 – 18

3. Draw the graph of y = x3.
Solution:
The given function is,

y = x3

x – 3 – 2 – 1 0 1 2 3
y – 27 – 8 – 1 0 1 8 27

PRIME Opt. Maths Book - X 91

4. Draw the graph of the function y = – 1 x3 .
2
Solution:

The given function is,

y = – 1 x3 .
2

x – 3 –2 – 1 0 1 2 3

y 13.5 4 0.5 0 – 0.5 – 4 – 13.5

5. Draw the graph of function y = 2x2 + 5x – 3.

Solution:
The given equation (quadratic) is,
y = 2x2 + 5x – 3
or, y = 2.x2 + 5x – 3

Here,

Vertex = a– b 4ac – b2 k
2a , 4a

=

= (– 1.25, – 6.125 )

Then, y = 2x2 + 5x – 3

x – 4 – 3 –2 – 1 0 1 2

y 9 0 – 5 – 6 –3 4 15

92 PRIME Opt. Maths Book - X

6. Solve the quadratic equation x2 – 5x + 4 = 0 graphically.

Solution:

The given quadratic equation is,
y = x2 – 5x + 4
or, y = 1.x2 – 5.x + 4

Here,

Vertex = a– b 4ac – b2 k
2a , 4a

=

= (2.5, – 2.25)

Then,
y = x2 – 5x + 4

x – 1 0 1 2 3 4 5 6
y 10 4 0 – 2 – 2 0 4 10

Here, we get from graph,
The curve of parabola cuts x - axis at point
A(4, 0) and B(1, 0).
\ x = 1 and 4

Alternative method:

The given quadratic equation is,
y = x2 – 5x + 4
Let, y = x2 ........................(i)
& y – 5x – 4 = 0
or, y = 5x – 4 .....................(ii)
For equation (i), y = x2,
y = 1.x2 + 0.x + 0 b2

\ vertex = a– b , 4ac – k = (0, 0)
2a 4a

x – 3 – 2 – 1 0 1 2 3 4
y 9 4 1 0 1 4 9 16

For equation (ii), y = 5x – 4

x012

y – 4 1 6

Here,
From graph, we get, the straight line cuts the parabola
at points A(1, 1) &
B(4, 16).
\ x = 1 & 4

PRIME Opt. Maths Book - X 93

Exercise 1.5 'A'

1. Answer the following questions.
i) What is quadratic equation? Write down the standard form of the quadratic

equation.

ii) Write down the co-ordinate of vertex of parabola.
iii) Write down the equation of axis of parabola.
iv) What do you mean by axis of parabola.
v) Write down the co-ordinate of the points on the curve of parabola at which x-axis

intersects.

2. Draw the graph of the following functions:

a) i) y = x2 ii) y = – x2 iii) y = 2x2

iv) y = – 3x2 v) y = 1 x2
2

b) i) Y = x3 ii) y = – x3 iii) y = 1 x3
iv) y = – 2x3 v) y = 2x3 2

c) i) y = x2 + x – 2 ii) y = x2 + 6x + 8 iii) y = x2 – 6x + 5
iv) y = x2 – 4 v) y = – x2 – x + 6

3. Solve the following quadratic equations graphically.

a) i) x2 + 2x – 3 = 0 ii) x2 + 5x + 4 = 0 iii) x2 – 6x +8 = 0
iii) 2x2 + 5x – 3 = 0
iv) x2 – x – 6 = 0 v) x2 – x – 12 = 0

b) i) – x2 – x + 6 = 0 ii) – x2 – 6x – 5 = 0
iv) 2x2 – 7x – 4 = 0 v) – 2x2 + 7x – 3 = 0

4. Solve the given functions graphically. iii) y = –x2 and y = – 3x – 4

i) y = x2 and y = 4x + 5 ii) y = x2 and y = 4x – 3
v) y = 2x2 and y = 2 – 3x v) y = x2 and y = 5x – 4

5. PRIME more creative questions.
i) A quadratic equation passer through the points (1, 1) and (2, 4) has the turning

point of (0, 0), find the equation of the quadratic function.
ii) A quadratic function passes through the points (0, 2), (1, –2) and (2, 0), find the

equation of the function.

94 PRIME Opt. Maths Book - X

iii) Find the equation of parabola having vertex (1, –4) which passer through the points
(2, –3) and (–1, 0).

iv) Find the equation of parabola from the given diagram.

(–2, 4)

(1, 1)

v) Find the equation of parabola from the given diagram.

(0, 3)
(1, 0)
(2, –1)

Answer

1. Show to your teacher.

2. Discuss with your teacher.

3.a. i) 1, – 3 ii) –1, –4 iii) –2, 4
iv) –2, 3 v) –3, 4
1
b. i) 2, –3 ii) –1, –5 iii) –3, 2
1 1
iv) 4, – 2 v) 3, 2

4. i) –1, 5 ii) 1, 3 iii) –1, 4
iv) –1, 4 v) 1, 4

5. i) y = x2 ii) y = 3x2 – 7x + 2 iii) y = x2 – 2x – 3
iv) y = x2 v) y = x2 – 4x + 3

PRIME Opt. Maths Book - X 95