Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{In an A.P., the series containing 99 terms, the sum of all}& \text{(P)}& 5010\\ & \text{the odd numbered terms is}2550.\text{The sum of all the}99\text{terms}& & \\ & \text{of the A.P. is}& & \\ \text{(B)}& f\text{is a function for which}f\left(1\right)=1\text{and}f\left(n\right)=n+f(n-1)& \text{(Q)}& 5049\\ & \text{for each natural number}n\ge 2\text{. The value of}f\left(100\right)\text{is}& & \\ \text{(C)}& \text{Suppose}f\left(n\right)={\log }_2\left(3\right)\cdot {\log }_3\left(4\right)\cdot {\log }_4\left(5\right)\dots {\log }_{n-1}\left(n\right).& \text{(R)}& 5050\\ & \text{Then the sum}\sum _{k=2}^{100}f\left(2^k\right)\text{equals}& & \\ \text{(D)}& \text{Concentric circles of radii}1,2,3,\dots ,100\text{cms}\text{are drawn. The}& \text{(S)}& 5100\\ & \text{interior of the smallest circle is coloured red and the annular}& & \\ & \text{regions are coloured alternately green and red, so that no}& & \\ & \text{two adjacent regions are of the same colour. The total area}& & \\ & \text{of the green regions in sq. cm is}k\pi .\text{Then}k\text{equals}& & \\ & & \text{(T)}& 5030\end{array}$

Which of the following is the only CORRECT combination?

• A. $\text{(C)}\to \text{(Q)},\text{(D)}\to \text{(R)}$
• B. $\text{(C)}\to \text{(Q)},\text{(D)}\to \text{(P)}$
• C. $\text{(C)}\to \text{(R)},\text{(D)}\to \text{(P)}$
• D. $\text{(C)}\to \text{(S)},\text{(D)}\to \text{(R)}$

Answer 1

The correct option is A $\text{(C)}\to \text{(Q)},\text{(D)}\to \text{(R)}$

$\left(C\right)$
$f\left(n\right)={\log }_2\left(3\right)\cdot {\log }_3\left(4\right)\cdot {\log }_4\left(5\right)\dots {\log }_{n-1}\left(n\right)$
$\Rightarrow f\left(n\right)=\frac{\log 3}{\log 2}\cdot \frac{\log 4}{\log 3}\cdots \frac{\log n}{n-1}$
$\Rightarrow f\left(n\right)=\frac{\log n}{\log 2}={\log }_2\left(n\right)$

$\therefore f\left(2^k\right)={\log }_2\left(2^k\right)=k$
$\therefore \sum _{k=2}^{100}f\left(2^k\right)$
$=\sum _{k=2}^{100}k=2+3+4+\cdots +100$
$=5050-1=5049$

$\left(D\right)$
Area $=\pi [(r_2^2-r_1^2)+(r_4^2-r_3^2)+\cdots +(r_{100}^2-r_{99}^2)]$
$r_2-r_1=r_4-r_3=\cdots =r_{100}-r_{99}=1$

$\therefore$ Area $=\pi [r_1+r_2+r_3+r_4+\cdots +r_{100}]$ $=\pi [1+2+3+\cdots +100]$ $=5050\pi \text{sq. cm}$