**For previous exercise – Progressions – Exercise 2.1 – Class X**

**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

**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 T_{1} = 2(1) – 1 = 2 – 1 = 1

(b) For n = 2 and T_{2} = 2(2) – 1 = 4 – 1 = 3

(c) For n = 3 and T_{3} = 2(3) – 1 = 6 – 1 = 5

(d) For n = 4 and T_{4} = 2(4) – 1 = 8 – 1 = 7

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

(ii) T_{n} = 5n + 1

(i) For n = 1 and T_{1} = 5(1) + 1 = 5 + 1 = 6

(ii) For n = 2 and T_{2} = 5(2) + 1 = 10 + 1 = 11

(iii) For n = 3 and T_{3} = 5(3) + 1 = 15 + 1 = 16

(iv) For n = 4 and T_{4} = 5(4) + 1 = 20 + 1 = 21

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

**In an A.P.**

**(i) if a = – 7, d = 5 find T _{12}**

**(ii) if a = – 1, d = -3 find T _{50}**

**(iii) if a = 12, d = 4 and T _{n} = 76, find n**

**(iv)if d = -2, T _{12} = -39, find a**

**(v) if a = 13, T _{15} = 55 , find d**

Solution:

For A.P. T_{n} = a + (n – 1)d

(i) if a = – 7, d = 5 , we have find T_{12}

T_{12} = a + (12 – 1)d

T_{12} = -7 + 11(5)

= 48

(ii) if a = – 1, d = -3, we have to find T_{50}

T_{50} = 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, T_{12} = -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, T_{15} = 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

**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

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

Solution:

T_{1} + T_{2} + T_{3} = 180˚

T_{2} = T_{1} + d

T_{3} = T_{1} + 2d

T_{1} + (T_{1} + d) + (T_{1} + 2d) = 180˚

3T_{1} + 3d = 180˚

T_{1} + d = 60˚

d = 60˚ – T_{1} = 60˚ – 50˚ = 10˚

T_{2} = T_{1} + d = 50˚ + 10˚ = 60˚

T_{3} = T_{1} + 2d = 50˚ + 2×10˚ = 50˚ + 20˚ = 70˚

**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, T_{3} = 12 and T_{50 }= 106. We have to find T_{29}

To find T_{29, }we need to find a and d,

d = ^{Tp – Tq}/_{p – q} = ^{106 – 12}/_{50 – 3} = 2

To find a,

T_{3} = a + (n – 1)d = a + (3 – 1)2 = a + 4

12 = a + 4

a = 12 – 4 = 8

To find T_{29},

T_{29} = a + (n – 1)d

= 8 + (29 – 1)2

T_{29 }= 64

**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

a_{4} = a +(4 -1)d

a_{4} = a + 3d

Similarly ,

a_{8} = a + 7d

a_{6} = a + 5d

a_{10} = a + 9d

Sum of 4^{th} and 8^{th} terms of an A.P = 24

a_{4} +a_{8} = 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

a_{6} +a_{10} = 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

a_{2} = a + d = -13 + 5 = -8

a_{3} = a_{2} + d = -8 + 5 = -3

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

**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}

**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.

**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.

**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 T_{1}, T_{2}, T_{3} and T_{4} i.e, a – 3d, a – d, a + d, a + 3d

T_{2} + T_{3} = 22

a + d + a – d = 22

2a = 22

a = 11

T_{1} x T_{4} = 85

(a – 3d) x (a + 3d) = 85

a^{2} – (3d)^{2} = 85

11^{2} – (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