Chapter 9: Sequences And Series Miscellaneous Exercise

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

We know that the k^th term of an A. P. is given by

a_k = a + (k –1) d

So, a_m + n = a + (m + n –1) d

And, a_m – n = a + (m – n –1) d

a_m = a + (m –1) d

Thus,

a_m + n + a_m – n  = a + (m + n –1) d + a + (m – n –1) d

= 2a + (m + n –1 + m – n –1) d

= 2a + (2m – 2) d

= 2a + 2 (m – 1) d

=2 [a + (m – 1) d]

= 2a_m

Therefore, the sum of (m + n)^th and (m – n)^th terms of an A.P. is equal to twice the m^th term

Let’s consider the three numbers in A.P. as a – d, a, and a + d.

Then, from the question we have

(a – d) + (a) + (a + d) = 24 … (i)

3a = 24

∴ a = 8

And,

(a – d) a (a + d) = 440 … (ii)

(8 – d) (8) (8 + d) = 440

(8 – d) (8 + d) = 55

64 – d² = 55

d² = 64 – 55 = 9

∴ d = ± 3

Thus,

When d = 3, the numbers are 5, 8, and 11 and

When d = –3, the numbers are 11, 8, and 5.

Therefore, the three numbers are 5, 8, and 11.

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

So, we have

First, let’s find the numbers between 200 and 400, which are divisible by 7.

The numbers are:

203, 210, 217, … 399

Here, the first term, a = 203

Last term, l = 399 and

Common difference, d = 7

Let’s consider the number of terms of the A.P. to be n.

Hence, a_n = 399 = a + (n –1) d

399 = 203 + (n –1) 7

7 (n –1) = 196

n –1 = 28

n = 29

Then, the sum of 29 terms of the A.P is given by:

Therefore, the required sum is 8729.

First, let’s find the integers from 1 to 100, which are divisible by 2.

And they are 2, 4, 6… 100.

Clearly, this forms an A.P. with the first term and common difference both equal to 2.

So, we have

100 = 2 + (n –1) 2

n = 50

Hence, the sum is

Now, the integers from 1 to 100, which are divisible by 5, are 5, 10… 100.

This also forms an A.P. with the first term and common difference both equal to 5.

So, we have

100 = 5 + (n –1) 5

5n = 100

n = 20

Hence, the sum is

Lastly, the integers which are divisible by both 2 and 5, are 10, 20, … 100.

And this also forms an A.P. with the first term and common difference both equal to 10.

So, we have

100 = 10 + (n –1) (10)

100 = 10n

n = 10

Thus, the required sum = 2550 + 1050 – 550 = 3050

Therefore, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

First, we have to find the two-digit numbers, which when divided by 4, yield 1 as the remainder.

They are: 13, 17, … 97.

As it’s seen that this series forms an A.P. with the first term (a) 13 and common difference (d) 4.

Let n be the number of terms of the A.P.

We know that the n^th term of an A.P. is given by,

a_n = a + (n –1) d

So, 97 = 13 + (n –1) (4)

4 (n –1) = 84

n – 1 = 21

n = 22

Now, the sum of n terms of an A.P. is given by,

Therefore, the required sum is 1210.

Given that,

f (x + y) = f (x) × f (y) for all x, y ∈ N … (1)

f (1) = 3

Taking x = y = 1 in (1), we have

f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9

Similarly,

f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27

And, f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81

Thus, f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with the first term and common ratio both equal to 3.

We know that sum of terms in G.P is given by,

And it’s given that,

Hence, the sum of the terms of the function is 120.

Therefore, the value of n is 4.

Given that the sum of some terms in a G.P is 315.

Let the number of terms be n.

We know that the sum of terms is

Given that the first term a is 5 and the common ratio r is 2.

Hence, the last term of the G.P = 6^th term = ar^{6 – 1} = (5)(2)⁵ = (5)(32) = 160

Therefore, the last term of the G.P. is 160.

Let’s consider a and r to be the first term and the common ratio of the G.P., respectively.

Given, a = 1

a₃ = ar² = r²

a₅ = ar⁴ = r⁴

Then, from the question we have

r² + r⁴ = 90

r⁴ + r² – 90 = 0

Therefore, the common ratio of the G.P. is ±3.

Let’s consider the three numbers in G.P. to be as a, ar, and ar².

Then from the question, we have

a + ar + ar² = 56

a (1 + r + r²) = 56

 … (1)

Also, given

a – 1, ar – 7, ar² – 21 form an A.P.

So, (ar – 7) – (a – 1) = (ar² – 21) – (ar – 7)

ar – a – 6 = ar² – ar – 14

ar² – 2ar + a = 8

ar² – ar – ar + a = 8

a(r² + 1 – 2r) = 8

a (r – 1)² = 8 … (2)

7(r² – 2r + 1) = 1 + r + r²

7r² – 14 r + 7 – 1 – r – r² = 0

6r² – 15r + 6 = 0

6r² – 12r – 3r + 6 = 0

6r (r – 2) – 3 (r – 2) = 0

(6r – 3) (r – 2) = 0

r = 2, 1/2

When r = 2, a = 8

When r = ½, a = 32

Thus,

When r = 2, the three numbers in G.P. are 8, 16, and 32.

When r = 1/2, the three numbers in G.P. are 32, 16, and 8.

Therefore, in either case, the required three numbers are 8, 16, and 32.