9. Sequences and Series Class 11 mathematics Solutions English Medium-Miscellaneous Exercise on Chapter - 9 (Available)

Miscellaneous Exercise on Chapter - 9

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Q1. Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
Solution: 
Let a and d be the first term and the common difference of the A.P. respectively. It is known that the k^th term of an A. P. is given by
ak = a + (k –1) d
∴ am + n = a + (m + n –1) d
am – n = a + (m – n –1) d
am = a + (m –1) d
∴ am + n + am – 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]
= 2am
Thus, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

Q2. If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.
Solution: 
Let the three numbers in A.P. be a – d, a, and a + d.
According to the given information,
(a – d) + (a) + (a + d) = 24 … (1)
⇒ 3a = 24
∴ a = 8
(a – d) a (a + d) = 440 … (2)
⇒ (8 – d) (8) (8 + d) = 440
⇒ (8 – d) (8 + d) = 55
⇒ 64 – d2 = 55
⇒ d2 = 64 – 55 = 9
⇒ d = ± 3
Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5.
Thus, the three numbers are 5, 8, and 11.

Q3. Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S₃ =3 (S₂– S₁).
Solution: 
Let a and b be the first term and the common difference of the A.P. respectively. Therefore,

Q4. Find the sum of all numbers between 200 and 400 which are divisible by 7.
Solution: 
The numbers lying between 200 and 400, which are divisible by 7, are
203, 210, 217 … 399
∴ First term, a = 203
Last term, l = 399
Common difference, d = 7

Let the number of terms of the A.P. be n.
∴ an = 399 = a + (n –1) d
⇒ 399 = 203 + (n –1) 7
⇒ 7 (n –1) = 196
⇒ n –1 = 28
⇒ n = 29

Sum of 29 terms is 8729.

Q5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Solution: 

Integers from 1 to 100 which are divisible by 2 are 2, 4, 6, 8 ................. 100. 

which form A.P : 2, 4, 6, 8 ................. 100.

Case-I

a = 2, d = d₂ - d₁ ⇒ 4 - 2 = 2 , a_n = 100,

a_n = a + (n - 1)d​ 

100 = 2 + (n - 1)2 

100 - 2 = (n - 1)2 

98 = (n - 1)2

n - 1 = 98/2

n - 1 = 49 

n = 49 + 1 

n = 50 

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Case-II

The integers between 1 to 100 divisible by 5.

5, 10, 15 ..................... 100.

a = 5, d = 10 - 5 = 5,

a_n = 100 

a_n = a + (n - 1)d​ 

⇒100 = 5 + (n - 1)5

⇒100 - 5 =  (n - 1)5

⇒95/5 = n - 1

⇒19 = n - 1

⇒n = 19 + 1

⇒n = 20

Q6. Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Solution:

The two-digit numbers, which when divided by 4, yield 1 as remainder, are as

A.P: 13, 17, … 97.
This series forms an A.P. 

a = 13, d = 17-13 = 4 a_n = 97

a_n = a + (n –1) d
∴ 97 = 13 + (n –1) (4)
⇒ 4 (n –1) = 84
⇒ n – 1 = 21
⇒ n = 22
Sum of n terms of an A.P. is given by,

Q8. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Solution: 

It is given that a = 5, r = 2, S_n = 315

n = ? and a_n = ? 

Q9. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

Solution: 

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

T₁ = a = 1

T₂ = ar = 1×r = r 

T₃ = ar² = 1× r² = r²

T₄ = ar³ = 1× r³ = r³

T₅ = ar⁴ = 1× r⁴ = r⁴

Given: T₃ + T₅ = 90

⇒r² +  r⁴ = 90 

⇒  r⁴ + r² - 90 =0 

Let  r² = y then gives;

y² + y - 90 = 0 

y² + 10y - 9y - 90 = 0 

y(y + 10) -9(y + 10) = 0 

(y + 10)(y - 9) = 0 

y + 10 = 0, y - 9 = 0 

y = -10 (Not to be taken) , y = 9 

y = 9 ⇒ r² = 9 ⇒ r = √9 ⇒ r = ± 3

Therefore, common ratios are 3, - 3. 

Q10. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Solution:

Let the three numbers in G.P. be a, ar, and ar².
According to condition,
a + ar + ar² = 56   ……………….… (1)

Now, a – 1, ar – 7, ar² – 21 are in A.P 

therefore, 

a1 = a – 1,

a₂ = ar – 7,

a₃ = ar² – 21 

a₂ - a₁ = a₃ - a_​2
∴ (ar – 7) – (a – 1) = (ar² – 21) – (ar – 7)
⇒ ar – a – 6 = ar² – ar – 14
⇒ar² – 2ar + a = 8
⇒a – 2ar + ar²   = 8  ........................ (2) 

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Q11. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Solution: 

Let the G.P be T₁, T₂, T₃, T₄ ................. T_2n

It is given that there are even numbers of terms. 

numbers of terms = 2n 

S = T₁ + T₂ + T₃ + T₄ + .................+ T_2n

S = a + ar + ar² + ……………….

First term = a, common ratio = r 

Let be sum of terms occupying odd places

S₁ = T₁ + T₃ + T₅ + .................+ T_2n-1

S₁ = a + ar² + ar⁴ + ……………….

First term = a, common ratio = r²

Total terms in this series = n  (half of 2n) 

Q12. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Solution: 

S₄ = 56, a = 11, Let total terms in A.P be n 

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