Sequence and Series Questions and Answers Part-4

1. If the ratio of sums to n terms of two A.P’s is (5n + 3) :(3n+4) then the ratio of their 17th terms is
a) 172 : 99
b) 168 : 103
c) 175 : 99
d) 171 : 103

Explanation: Let two A.P.’s be

2. If $H_{1},H_{2},....,H_{n}$   are n harmonic means between a and $b\left(\neq a\right)$  , then value of $\frac{H_{1}+a}{H_{1}-a}+\frac{H_{n}+b}{H_{n}-b}$    is equal to
a) n + 1
b) n - 1
c) 2n
d) 2n + 3

Explanation: As a, H1, H2, .... , Hn, b are in H.P

3. Let $a_{1},a_{2},....,a_{n}$    be n positive real numbers such that $a_{1}a_{2}....a_{n}=c$    , where c number. The expression $a_{1}+2a_{2}+3a_{3}+....+na_{n}$     cannot be less than
a) n! c
b) $n\left(n! c\right)^{1/n}$
c) $C^{1/n}$
d) $\left(2n\right)C^{1/n}$

Explanation:

4. The sum upto (2n + 1) terms of the series $a^{2}-\left(a+d\right)^{2}+\left(a+2d\right)^{2}-\left(a+3d\right)^{2}+....$
is
a) $a^{2}+3nd ^{2}$
b) $a^{2}+2nad +n\left(n-1\right)d^{2}$
c) $a^{2}+3nad +n\left(n-1\right)d^{2}$
d) $a^{2}+2nad +n\left(2n+1\right)d^{2}$

Explanation: We can write the sum upto (2n + 1) terms as

5. Value of y=$(0.64)^{\log_{0.25}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+.... \infty\right)}$
is
a) 0.9
b) 0.8
c) 0.6
d) 0.25

Explanation:

6. Let $ax^{2}+\frac{b}{x}\geq c$   for all positive x, where a > 0 and b>0. The value of the expression $27 ab^{2}$ cannot be less than
a) $4c^{3}$
b) $4c^{2}$
c) $8c^{3}$
d) $c^{3}$

Explanation: As A.M. ≥ G.M., we get

7.Let $b_{i}>1$  for i = 1, 2, …, 101. Suppose $\log_{e}b_{1},\log_{e}b_{2},.....,\log_{e}b_{101}$      are in Arithmetic Progression (A.P.) with the common difference $log_{e} 2$ . Suppose $a_{1},a_{2},....,a_{101}$    are in A.P. such that $a_{1}=b_{1}$  and $a_{51}=b_{51}$  . If $t= b_{1}+b_{2}+....+b_{51}$     and $S= a_{1}+a_{2}+....+a_{51}$    , then
a) $S> t$   and $a_{101}>b_{101}$
b) $S> t$   and $a_{101}< b_{101}$
c) $S< t$   and $a_{101}>b_{101}$
d) $S< t$   and $a_{101}< b_{101}$

Explanation: logeb1, logeb2, .…, logeb101 are in A.P. with

8.If $S_{n}=\sum_{r=1}^{n}t_{r}=\frac{1}{6}n\left(2n^{2}+9n+13\right)$      , then $\sum_{r=1}^{n}\sqrt{t_{r}}$   equals
a) $\frac{1}{2}n\left(n+1\right)$
b) $\frac{1}{2}n\left(n+2\right)$
c) $\frac{1}{2}n\left(n+3\right)$
d) $\frac{1}{2}n\left(n+5\right)$

Explanation:

9. If 0< $\theta$ , $\phi < \pi/2$   and $x=\sum_{n=0}^{\infty} \sin^{2n}\theta,y=\sum_{n=0}^{\infty} \cos^{2n}\phi,$
and $z=\sum_{n=0}^{\infty} \cos^{n}\left(\theta+\phi\right)\cos^{n}\left(\theta-\phi\right)$
a) xyz + 1 = yz – zx
b) xyz – 1 = yz + zx
c) xyz – xy = yz – zx
d) xyz + 1 = yz + zx

10. If $a_{1},a_{2},....,a_{n}$    are in A.P. with common difference d $\neq$ 0, then sum of the series $\sin d\left[\sec a_{1}\sec a_{2}+\sec a_{2}\sec a_{3}+....+\sec a_{n-1}\sec a_{n}\right]$
a) $\tan a_{n}-\tan a_{1}$
b) $\cot a_{n}-\cot a_{1}$
c) $\sec a_{n}-\sec a_{1}$