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 7 of 8 pages

Question 61. If a and b are two different positive real numbers, then which of the following relations is true

2√ab>(a+b)
2√ab
2√ab = (a+b)
None of these

Discuss Question

Answer: Option B. -> 2√ab
:
B
We know that A > G > H
Where A is arithmetic mean, G is geometric mean and H is harmonic mean, then A > G

\Rightarrow \frac{a + b}{2} > \sqrt{a b}

or (a + b) > 2

\sqrt{a b}

Question 62. Suppose that all the terms of an arithmetic progression are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this AP is___

Discuss Question

:
Given,

\frac{S_{7}}{S_{11}} = \frac{6}{11} a n d 130 < t_{7} < 140

\Rightarrow \frac{\frac{7}{2} [2 a + 6 d]}{\frac{11}{2} [2 a + 10 d]} = \frac{6}{11}

\Rightarrow \frac{7 (2 a + 6 d)}{(2 a + 10 d)} = 6

\Rightarrow a = 9 d \dots \dots (i)

Also,

130 < t_{7} < 140

\Rightarrow 130 < a + 6 d < 140

\Rightarrow 130 < 9 d + 6 d < 140

[from Eq. (i)]

\Rightarrow 130 < 15 d < 140

\Rightarrow \frac{26}{3} < d < \frac{28}{3}

∴

d=9 [since, d is a natural number]

Question 63. If

a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}}

and

a, b, c

are in G.P., then which of the following are true?

x,y,z will be in A.P.
x,y,z will be in G.P.
x,y,z will be in H.P.
None of the above

Discuss Question

Answer: Option A. -> x,y,z will be in A.P.
:
A
Let

a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} = k .

\Rightarrow a = k^{x}, b = k^{y}, c = k^{z} .

Since

a, b, c

are in G.P.,

b^{2} = a c .

∴ k^{2 y} = k^{x} . k^{z} = k^{x + z}

∴ 2 y = x + z

\Rightarrow x, y, z

are in A.P.

Question 64. The sum of n terms of the series whose

n^{t h}

term is n(n+1) is equal to

n(n+1)(n+2)3
(n+1)(n+2)(n+3)12
n2(n+2)
n(n+1)(n+2)

Discuss Question

Answer: Option A. -> n(n+1)(n+2)3
:
A

T_{π}

n^{2}

+ n ⇒

S_{π}

\sum T_{π}

\sum

n^{2}

\sum

n
=

\frac{n (n + 1) (2 n + 1)}{6}

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

\frac{n (n + 1)}{6}

{2n + 1 + 3} =

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

Question 65. The sum of the series

1^{2}

.2 +

2^{2}

.3 +

3^{2}

.4 + ........ to n terms is

n3(n+1)3(2n+1)24
n(n+1)(3n2+7n+2)12
n(n+1)6[n(n+1) + (2n + 1)]
n(n+1)12[6n(n+1) + 2(2n + 1)]

Discuss Question

Answer: Option B. -> n(n+1)(3n2+7n+2)12
:
B

T_{π}

n^{2}

(n + 1) =

n^{3}

n^{2}

S_{π}

\sum

T_{π}

\sum

n^{3}

\sum

n^{2}

{[\frac{n (n + 1)}{2}]}^{2}

\frac{n (n + 1) (2 n + 1)}{6}

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

[

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

\frac{2 n + 1}{3}

] =

\frac{n (n + 1) (3 n^{2} + 7 n + 2)}{12}

Question 66. The terms of a G.P. are positive. If each term is equal to the sum of two terms that follow it, then the common ratio is ___.

√5−12
−1−√52
1
2√5

Discuss Question

Answer: Option A. -> √5−12
:
A
The given condition can be expressed as

T_{n}

T_{n + 1} + T_{n + 2},

where

n \geq 1.

a

is the first term and

r

is the common ratio, then by the condition is

a r^{n - 1} = a r^{n} + a r^{n + 1} . \Rightarrow r^{n - 1} = r^{n} (1 + r)

\Rightarrow r^{2} + r - 1 = 0

\Rightarrow r = \frac{- 1 \pm \sqrt{1 + 4}}{2} = \frac{- 1 \pm \sqrt{5}}{2}

Since each term is positive, the common ratio of the GP is

\frac{\sqrt{5} - 1}{2} .

Question 67. If the ratio of the sum of first three terms and the sum of first six terms of a G.P. be 125 : 152, then the common ratio

r

is ___.

Discuss Question

Answer: Option A. -> 35
:
A
Given that

\frac{a + a r + a r^{2}}{a + a r + a r^{2} + a r^{3} + a r^{4} + a r^{5}} = \frac{125}{152} .

\Rightarrow \frac{1 + r + r^{2}}{1 + r + r^{2} + r^{3} + r^{4} + r^{5}} = \frac{125}{152} i.e., \frac{1 + r + r^{2}}{(1 + r + r^{2}) + r^{3} (1 + r + r^{2})} = \frac{125}{152} \Rightarrow \frac{1}{1 + r^{3}} = \frac{125}{152} \Rightarrow r^{3} = \frac{152}{125} - 1 = \frac{27}{125} \Rightarrow r = \frac{3}{5}

Question 68. The number of terms between 1 to 1000 divisible by 7 are ___.

Discuss Question

Answer: Option A. -> 142
:
A
The required sequence will be 7,14, 21, 28, . . . . 994.
Now, the number of terms divisible by 7 will be
994 = 7 + (n - 1)7
[

∵ a_{n}

= a + (n - 1) d]
Where,
a = 7,
d = 14 -7 = 7 and
n = Number of terms.

\Rightarrow 994 = 7 + 7 n - 7

\Rightarrow 994 = 7 n

\Rightarrow n = \frac{994}{7}

\Rightarrow n = 142

Hence, the number of terms between 1 to 1000 that are divisible by 7 are 142.

Question 69. In a flag race, a pole is placed at the starting point, which is 10m from the first flag and the other flags are placed 6m apart in a straight line. There are 10 flags in the line. Each competitor starts from the pole, picks up the nearest flag and comes back to the pole and runs for the next flag and continues the same way until all the flags are on the pole. The total distance covered is ____.