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

---

2. Find the sequence if

(i)  T_n = 2n – 1

(ii) T_n = 5n – 1

Solution:

(i)  T_n = 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) T_n = 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,…

---

3. 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 T_n = 76, find n

(iv)if d = -2, T₁₂ = -39, find a

(v) if a = 13, T₁₅ = 55  , find d

Solution:

For A.P. T_n = 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 T_n = 76, we have to find n

T_n = 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, T_n = 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, T_n = a + (n – 1)d

55 = 13 + (15 – 1)d

55 – 13 = 14d

42 = 14d

d = ⁴²/₁₄ = 3

---

4. 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, T_n = a + (n – 1)d

100 = 12 + (n – 1)4

100 = 12 + 4n – 4

100 – 12 + 4 = 4n

92 = 4n

n = ⁹²/₄ = 23

---

5. 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˚

---

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

Solution:

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

To find T_29, 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

---

7. The sum of 4^th and 8^th terms of an A.P. is 24 and the sum of 6^th and 10^th 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
a_n = a+(n-1) d
a₄ = a +(4 -1)d
a₄ = a + 3d
Similarly ,
a₈ = a + 7d
a₆  = a + 5d
a₁₀ = a + 9d
Sum of 4^th and 8^th 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

---

8. The ratio of 7^th to 3^rd term of an A.P. is 12 : 5 . Find the ratio of 13^th to 4^th 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/_7­)

^T13/_T4 = ^90d/_27d

^T13/_T4 = ¹⁰/₃

---

9. 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.

---

10. 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:

T_p = q = a + (p – 1)d ……………(1)

T_q = 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, T_n = a + (n – 1)d

T_n = (q + p – 1) + (n – 1)(-1)

T_n = q + p – 1 – n + 1

T_n = q + p – n

Hence proved.

---

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

---