Sequences And Series(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

SEQUENCES AND SERIES MCQs

Sequence And Series

Total Questions : 75 | Page 6 of 8 pages

Question 51. 1 . The sum of the series 1 + 3x + 6

x^{2}

+ 10

x^{3}

+ ....... ∞ will be

1(1−x)2
11−x
1(1+x)2
1(1−x)3

Discuss Question

Answer: Option D. -> 1(1−x)3
:
D
Let S = 1 + 3x + 6

x^{2}

+ 10

x^{3}

+ .......∞
⇒ x.S = x + 3

x^{2}

+ 6

x^{3}

+ ..........∞
Subtracting S(1-x) = 1 + 2x + 3

x^{2}

+ 4

x^{3}

+ ........∞
⇒ x(1 - x)S = x + 2

x^{2}

+ 3

x^{3}

+ .........∞
Again subtracting,
⇒ S[(1 - x) - x(1 - x)] = 1 + x +

x^{2}

x^{3}

+ ...........∞
⇒ S[(1 - x)(1 - x)] =

\frac{1}{1 - x}

⇒ S =

\frac{1}{(1 - x)^{3}}

Question 52. If

\frac{1}{b - a} + \frac{1}{b - c}

\frac{1}{a} + \frac{1}{c}

, then a,b,c are in

A.P
G.P
H.P
In G.P and H.P both

Discuss Question

Answer: Option C. -> H.P
:
C
Since the reciprocals of a and c occur on RHS, let us first assume that a,b,c are in H.P.
So that

\frac{1}{a}

\frac{1}{b}

\frac{1}{c}

are in A.P.

\Rightarrow \frac{1}{b} - \frac{1}{a}

\frac{1}{c} - \frac{1}{b}

= d,say

\Rightarrow \frac{a - b}{a b}

= d =

\frac{b - c}{b c} \Rightarrow a - b

= abd and b - c =bcd
Now LHS =

- \frac{1}{a - b} + \frac{1}{b + c}

- \frac{1}{a b d} + \frac{1}{b c d}

= \frac{1}{b d} (\frac{1}{c} - \frac{1}{a})

\frac{1}{b d} (2 d) ∴ \frac{2}{b}

\frac{1}{a} + \frac{1}{c}

= RHS

∴

a, b, c are in H.P. is verified.

Question 53. A series in G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying odd places, then the common ratio will be equal to

Discuss Question

Answer: Option C. -> 4
:
C
Let there be 2n terms in the given G.P. with first term a and the common ratio r.
Then, a

\frac{r^{2} - 1}{(r - 1)}

= 5a

\frac{(r^{2} - 1)}{(r^{2} - 1)}

⇒ r + 1 = 5⇒ r = 4

Question 54. If the sum of the n terms of G.P. is S product is P and sum of their inverse is R, than

P^{2}

is equal to

RS
SR
(RS)n
(SR)n

Discuss Question

Answer: Option D. -> (SR)n
:
D

S = a \frac{(1 - r^{n})}{1 - r} P = a^{n} r \frac{n (n - 1)}{2} R = \frac{1}{a} + \frac{1}{a r} + \frac{1}{a r^{2}} + \dots \dots + \frac{1}{a r^{n - 1}} = \frac{\frac{1}{a} (1 - \frac{1}{r^{n}})}{1 - \frac{1}{r}} = \frac{1}{a r^{n - 1}} (\frac{r^{n} - 1}{r - 1}) p^{2} = a^{2 n} r^{n (n - 1)} (\frac{S}{R}) = a^{2 n} r^{n (n - 1)}

Question 55. The sum of (n+1) terms of

\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + . . . . . . . . . .

nn+1
2nn+1
2n(n+1)
2(n+1)n+2

Discuss Question

Answer: Option D. -> 2(n+1)n+2
:
D

T_{n} = \frac{1}{[\frac{n (n + 1)}{2}]} = 2 [\frac{1}{n} - \frac{1}{n + 1}]

put n=1,2,3,.........,(n+1)

T_{1} = 2 [\frac{1}{1} - \frac{1}{2}]

T_{2} = 2 [\frac{1}{2} - \frac{1}{3}]

,...........,

T_{n + 1} = 2 [\frac{1}{n + 1} - \frac{1}{n + 2}]

Hence sum of (n+1) terms =

n + 1 \sum k = 1 T_{k}

\Rightarrow

S_{n + 1} = 2 [1 - \frac{1}{n + 2}]

\Rightarrow

S_{n + 1} = \frac{2 (n + 1)}{n + 2}

Question 56.

\frac{1}{1.2}

\frac{1}{2.3}

\frac{1}{3.4}

+.......+ .........

\frac{1}{n . (n + 1)}

equals

1n(n+1)
nn+1
2nn+1
2n(n+1)

Discuss Question

Answer: Option B. -> nn+1
:
B
(

\frac{1}{1}

\frac{1}{2}

) + (

\frac{1}{2}

\frac{1}{3}

) + (

\frac{1}{3}

\frac{1}{4}

) + .........+ (

\frac{1}{n}

\frac{1}{n + 1}

)
= 1 -

\frac{1}{n + 1}

\frac{n}{n + 1}

Question 57. Let

S_{1}

S_{2}

,....... be squares such that for each n≥1, the length
of a side of

S_{n}

equals the length of a diagonal of

S_{n + 1}

. If the
length of a side of

S_{1}

is 10 cm, then for which of the following
values of n is the area of

S_{n}

greater then 1sq cm

Discuss Question

Answer: Option A. -> 7
:
A
(b, c, d) Given

x_{n}

x_{n + 1}

\sqrt{2}

∴

x_{1}

x_{2}

\sqrt{2}

x_{2}

x_{3}

\sqrt{2}

x_{n}

x_{n + 1}

\sqrt{2}

On multiplying

x_{1}

x_{n + 1}

{(\sqrt{2})}^{n}

⇒

x_{n + 1}

\frac{x_{1}}{(\sqrt{2})^{n}}

Hence

x_{n}

\frac{x_{1}}{(\sqrt{2})^{n - 1}}

Area of

S_{n}

x_{n}^{2}

\frac{x_{n}^{2}}{2^{n - 1}}

< 1 ⇒

2^{n - 1}

x_{1}^{2}

(

x_{1}

= 10)
∴

2^{n - 1}

> 100
But

2^{7}

> 100,

2^{8}

>100, etc.
∴ n - 1 = 7, 8, 9....... ⇒ n = 8, 9, 10.........

Question 58. If

(p + q)^{th}

term and

(p - q)^{th}

term of a G.P. be

m

and

n,

then the

p^{th}

term will be ___.

mn
√mn
mn
o

Discuss Question

Answer: Option B. -> √mn
:
B
Given that

a_{p + q} = a r^{p + q - 1} = m

and

a_{p - q} = a r^{p - q - 1} = n .

\Rightarrow m \times n = a r^{p + q - 1}

a r^{p - q - 1}

= a^{2} r^{2 (p - 1)}

= (a r^{p - 1})^{2}

\Rightarrow \sqrt{m n} = a r^{p - 1} = a_{p}

Thus, the

p^{th}

term of the GP is

\sqrt{m n} .

Aliter: Each term in a G.P. is the geometric mean of the terms equidistant from it. Here,

(p + q)^{t h}

and

(p - q)^{t h}

terms are equidistant from the

p^{t h}

term.

∴

p^{t h}

term(

\sqrt{m n}

) will be G.M. of

(p + q)^{t h}

and

(p - q)^{t h}

terms.

Question 59. If

π \sum i = 1 i

\frac{n (n + 1)}{2}

, then

π \sum i = 1 (3 i - 2)

n(3n−1)2
n(3n+1)2
n(3n + 2)
n(3n+1)4

Discuss Question

Answer: Option A. -> n(3n−1)2
:
A

π \sum i = 1

= 3

π \sum i = 1 i

- 2

π \sum i = 1 1

= 3

\frac{n (n + 1)}{2}

- 2n =

\frac{n (3 n - 1)}{2}

Question 60. If the sum of two extreme numbers of an A.P. with four terms is 8 and product of remaining two middle term is 15, then greatest number of the series will be

Discuss Question

Answer: Option B. -> 7
:
B
Let four numbers are