Question

Match List-I with List-II.

	List-I		List-II
P.	If $t_1,t_2,t_3,t_4,t_5$ are first five terms of an A.P. $\frac{4(t_1-t_2-t_4)+6t_3+t_5}{t_1}$ then is equal to $(t_1\ne 0)$	(1)	2
Q.	$\sum _{K=1}^{\infty }\sum _{n=1}^{\infty }\frac{K}{2^{n+K}}$is less than or equal to	(2)	3
R.	If $T_n$ denotes the nth term of the H.P. from the beginning such that $\frac{T_2}{T_6}=9,$ then the value of $\frac{T_5}{T_{12}}$ is divisible by	(3)	4
S.	The value of $\frac{2}{1^3}+\frac{6}{1^3+2^3}+\frac{12}{1^3+2^3+3^3}$$+\frac{20}{1^3+2^3+3^3+4^3}+..$ upto infinity is less than or equal to	(4)	5
		(5)	6
		(6)	1

• A. P(2); Q(1, 2, 3, 4, 5); R(2, 6); S(3, 4, 5)
• B. P(1); Q(2, 6); R(3, 4); S(4, 5)
• C. P(2); Q(3, 4, 5); R(1, 2, 3, 4, 5); S(2, 6)
• D. P(1); Q(2); R(2); S(3)

Answer 1

The correct option is A P(2); Q(1, 2, 3, 4, 5); R(2, 6); S(3, 4, 5)

(P) $\frac{4\left(a(a+d)(a+3d)\right)+6(a+2d)+(a+4d)}{a}$
$=\frac{3a}{a}=3$
(Q) $\sum _{k=1}^{\infty }(\frac{k}{2^{1+k}}+\frac{k}{2^{2+k}}+\frac{k}{2^{3+k}}+...\mathrm{upto}\infty )$
$=\sum _{k=1}^{\infty }\frac{k}{\frac{2^{1+k}}{1\frac{1}{2}}}=\sum _{k=1}^{\infty }\frac{k}{2^k}$
Let $\begin{array}{cc}& \underset{–}{\begin{array}{cc}& S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+.....\infty \\ & \frac{S}{2}=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+......\infty \end{array}}\\ & \frac{S}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....\end{array}$
$\Rightarrow S=2$

(R) $\frac{T_2}{T_6}=9$

$\Rightarrow$ $\frac{\frac{1}{a+d}}{\frac{1}{a+5d}}=9$

$\Rightarrow$ $\frac{a+5d}{a+d}=9$

$\Rightarrow$ $9a+9d=a+5d$
$\Rightarrow$ $d=–2a$
$\therefore \frac{T_5}{T_{12}}=\frac{\frac{1}{a+4d}}{\frac{1}{a+11d}}=\frac{a+11d}{a+4d}=\frac{a-22a}{a-8a}=\frac{-21a}{-7a}=3$

(S) $T_n=\frac{n(n+1)}{{\left(\frac{n(n+1)}{2}\right)}^2}=\frac{4}{n(n+1)}$

$\to$ $T_n=4\left(\frac{1}{n}\frac{1}{n+1}\right)$

$\to$ $S_n=4(1-\frac{1}{n+1})$
${lim}_{n\to \infty }{S}_n=4$