Chapter 9: Sequences and Series Exercise 9.2

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

It clearly forms a sequence in A.P.

Where, the first term, a = 1

The common difference, d = 2

Now,

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n – 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

S_n = n/2 [2a + (n-1)d]

Therefore, the sum of odd numbers from 1 to 2001 is 1002001.

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

It clearly forms a sequence in A.P.

Where, the first term, a = 105

The common difference, d = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

105 + 5n – 5 = 995

5n = 995 – 105 + 5 = 895

n = 895/5

n = 179

We know,

S_n = n/2 [2a + (n-1)d]

Therefore, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

Given,

The first term (a) of an A.P. = 2

Let’s assume d be the common difference of the A.P.

So, the A.P. will be 2, 2 + d, 2 + 2d, 2 + 3d, …

Then,

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

From the question, we have

10 + 10d = ¼ (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a₂₀ = a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112

Therefore, the 20^th term of the A.P. is –112.

Let’s consider the sum of n terms of the given A.P. as –25.

We know that,

S_n = n/2 [2a + (n-1)d]

where n = number of terms, a = first term, and d = common difference

So here, a = –6

d = -11/2 + 6 = (-11 + 12)/ 2 = 1/2

Thus, we have

Given A.P.,

25, 22, 19, …

Here,

First term, a = 25 and

Common difference, d = 22 – 25 = -3

Also given, the sum of a certain number of terms of the A.P. is 116

The number of terms be n

So, we have

S_n = n/2 [2a + (n-1)d] = 116

116 = n/2 [2(25) + (n-1)(-3)]

116 x 2 = n [50 – 3n + 3]

232 = n [53 – 3n]

232 = 53n – 3n²

3n² – 53n + 232 = 0

3n² – 24n – 29n+ 232 = 0

3n(n – 8) – 29(n – 8) = 0

(3n – 29) (n – 8) = 0

Hence,

n = 29/3 or n = 8

As n can only be an integral value, n = 8

Thus, the 8^th term is the last term of the A.P.

a₈ = 25 + (8 – 1)(-3)

= 25 – 21

= 4

Given, the k^th term of the A.P. is 5k + 1.

k^th term = a_k = a + (k – 1)d

And,

a + (k – 1)d = 5k + 1

a + kd – d = 5k + 1

On comparing the coefficient of k, we get d = 5

a – d = 1

a – 5 = 1

⇒ a = 6

We know that,

S_n = n/2 [2a + (n-1)d]

From the question we have,

On comparing the coefficients of n² on both sides, we get

d/2 = q

Hence, d = 2q

Therefore, the common difference of the A.P. is 2q.

Let a₁, a₂, and d₁, d₂ be the first terms and the common difference of the first and second arithmetic progression, respectively.

Then, from the question we have

Let’s take a and d to be the first term and the common difference of the A.P., respectively.

Then, it is given that

Therefore, the sum of (p + q) terms of the A.P. is 0.