# Progressions – Exercise 2.2 – Class X

For previous exercise – Progressions – Exercise 2.1 – Class X

1. Write first four terms of the following A.P

(i) 0, -3, -6, -9,

(ii) 1/6, 1/3, 1/2,

(iii) a + b, a – b, a – 3b

Solution:

(i)  -12, -15, -18,  -21

(ii) 2/3, 5/6, 1, 7/6,

(iii) a – 5b, a – 7b, a – 9b, a – 11b

1. Find the sequence if

(i)  Tn = 2n – 1

(ii) Tn = 5n – 1

Solution:

(i)  Tn = 2n – 1

(a) For n = 1 and T1 = 2(1) – 1 = 2 – 1 = 1

(b) For n = 2 and T2 = 2(2) – 1 = 4 – 1 = 3

(c) For n = 3 and T3 = 2(3) – 1 = 6 – 1 = 5

(d) For n = 4 and T4 = 2(4) – 1 = 8 – 1 = 7

Therefore, the sequence is 1, 3, 5, 7, ….

(ii) Tn = 5n + 1

(i) For n = 1 and T1 = 5(1) + 1  = 5 + 1 = 6

(ii) For n = 2 and T2 = 5(2) + 1  = 10 + 1 = 11

(iii) For n = 3 and T3 = 5(3) + 1  = 15 + 1 = 16

(iv) For n = 4 and T4 = 5(4) + 1  = 20 + 1 = 21

Therefore, the sequence is 6, 11, 16, 21,…

1. In an A.P.

(i) if a = – 7, d = 5 find T12

(ii) if a = – 1, d = -3 find T50

(iii)  if a = 12, d = 4 and Tn = 76, find n

(iv)if d = -2, T12 = -39, find a

(v) if a = 13, T15 = 55  , find d

Solution:

For A.P. Tn = a + (n – 1)d

(i) if a = – 7, d = 5 , we have find T12

T12 = a + (12 – 1)d

T12 = -7 + 11(5)

= 48

(ii) if a = – 1, d = -3, we have to find T50

T50 = a + (50 – 1)d

= a + 49d

= -1 + 49(-3)

= – 148

(iii)  if a = 12, d = 4 and Tn = 76, we have to find n

Tn = a + (n – 1)d

76 = 12 + (n – 1)4

76 = 12 + 4n – 4

76 + 4 – 12 = 4n

68 = 4n

n = 17

(iv) if d = -2, T12 = -39, we have to find a

We know, Tn = a + (n – 1)d

-39 = a + (12 – 1)(-2)

– 39 = a – 24  + 2

a = – 39 +  22

a = -17

(v) if a = 13, T15 = 55  , we have to find d

We know, Tn = a + (n – 1)d

55 = 13 + (15 – 1)d

55 – 13 = 14d

42 = 14d

d = 42/14 = 3

1. Find the number of terms in the A.P., 100, 96, 92 …….12

Solution:

a = 12

d = 100 – 96 = 4

Tn = 100

We have find n,

We know, Tn = a + (n – 1)d

100 = 12 + (n – 1)4

100 = 12 + 4n – 4

100 – 12 + 4 = 4n

92 = 4n

n = 92/4 = 23

1. The angles of a triangle are in A.P. If the smallest angle is 50˚, find the other two angles.

Solution:

T1 + T2 + T3 = 180˚

T2 = T1 + d

T3 = T1 + 2d

T1 + (T1 + d) + (T1 + 2d) = 180˚

3T1 + 3d = 180˚

T1 + d = 60˚

d = 60˚ – T1 = 60˚ – 50˚ = 10˚

T2 = T1 + d = 50˚ + 10˚ = 60˚

T3 = T1 + 2d = 50˚ + 2×10˚ = 50˚ + 20˚ = 70˚

1. An A.P. consists of 50 terms of which 3rd term is 12 and last term is 106. Find the 29th term

Solution:

Given, T3 = 12 and T50 = 106. We have to find T29

To find T29, we need to find a and d,

d = Tp – Tq/p – q = 106 – 12/50 – 3 = 2

To find a,

T3 = a + (n – 1)d = a + (3 – 1)2 = a + 4

12 = a + 4

a = 12 – 4 = 8

To find T29,

T29 = a + (n – 1)d

= 8 + (29 – 1)2

T29 = 64

1. The sum of 4th and 8th terms of an A.P. is 24 and the sum of 6th and 10th terms of the same A.P. is 44. Find the first three terms

Solution:

Let the first term of an A.P = a
and the common difference of the given  A.P = d
As we know that
an = a+(n-1) d
a4 = a +(4 -1)d
a4 = a + 3d
Similarly ,
a8 = a + 7d
a6  = a + 5d
a10 = a + 9d
Sum of 4th and 8th terms of an A.P = 24
a4 +a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12  …………………(i)
Sum of 6 th and 10 th term  of an A.P = 44
a6 +a10 = 44
a + 5d + a + 9d = 44
2a + 14 =44
a + 7d = 22  …………………(ii)
From (i) & (ii)
a +7d  = 22
a + 5d = 12
–   –       –

————————

2d  = 10

d = 5

From equation (i) ,
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = – 13
a2 = a + d = -13 + 5 = -8
a3 = a2 + d = -8 + 5 = -3

The first three terms are -13 , -8, -3

1. The ratio of 7th to 3rd term of an A.P. is 12 : 5 . Find the ratio of 13th to 4th term

Solution:

T7/T3 = 12/5

a+6d/a+2d = 12/5

12(a + 2d) = 5(a + 6d)

12a + 24d = 5a + 30d

12a – 5a = 30d – 24d

7a = 6d

a = 6d/7

Then, T13/T4 = a+12d/a+3d = (6d/7 + 12d)/( 6d/7 + 3d)

T13/T4 = (6d+84d/7)/(6d+21d/7)

T13/T4 = (90d/7)/(27d/)

T13/T4 = 90d/27d

T13/T4 = 10/3

1. A company employed 400 persons in the year 2001 and each year increased by 35 persons. In which year the number of employees in the company will be 785?

Solution:

a = 400, d = 35, Tn = 785 So, we have to find n.

Tn = a + (n – 1)d

785 = 400 + (n – 1)35

785 = 400 + 35n – 35

785 + 35 – 400 = 35n

420 = 35n

n = 420/35 = 12

In 12 years the number of employees in the company will be 785 i.e., in the year 2013 the number of employees in the company will be 785.

1. If the pth term of an A.P. is q and the qth term is p, prove that the nth term is equal to (p + q – n)

Solution:

Tp = q = a + (p – 1)d ……………(1)

Tq = p = a + (q – 1)d ………………(2)

By subtracting (1) from (2), we get,

(p – 1)d – (q – 1)d = q – p

d[p – 1 – q + 1] = q – p

d[ p – q ] = q – p

d = q – p/p – q = – 1…………..(3)

Put the value of d in (1),

q = a + (p – 1)(-1)

q = a – p + 1

a = q + p – 1 ……………..(4)

We know that, Tn = a + (n – 1)d

Tn = (q + p – 1) + (n – 1)(-1)

Tn = q + p – 1 – n + 1

Tn = q + p – n

Hence proved.

1. Find four numbers in A.P. such that the sum of 2nd nd 3rd terms in 22 and product of 1st and 4th terms is 85?

Solution:

Let the four terms be T1, T2, T3 and T4 i.e, a – 3d, a – d, a + d, a + 3d

T2 + T3 = 22

a + d + a – d = 22

2a  = 22

a = 11

T1 x T4 = 85

(a – 3d) x (a + 3d) = 85

a2 – (3d)2 = 85

112 – (3d)2 = 85

121 – (3d)2 = 85

(3d)2 = – 85 + 121

(3d)2 = 36

3d = 6

d = 6/3 = 2

Therefore, a – 3d = 11 – 3(2) = 11 – 6 = 5

a – d = 11 – 2 = 9

a + d = 11 + 2 = 13

a + 3d = 11 + 3(2) = 11 + 6 = 17

Thus, four terms are 5, 9, 13, 17 or 17, 13, 9, 5