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

Answer 1

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

$\left(A\right)$
A.P. is $a,(a+d),(a+2d),\dots ,(a+98d)$
Sum of odd terms $=2550$
$\underset{50\text{terms}}{\underset{}{a+(a+2d)+(a+4d)+\cdots +(a+98d)}}=2550$
$\Rightarrow \frac{50}{2}[2a+98d]=2550$
$\Rightarrow 50[a+49d]$ $=2550$
$\Rightarrow a+49d=51$
Hence, $S_{99}=\frac{99}{2}[2a+98d]=99[a+49d]$
$=51\times 99=5049$

$\left(B\right)$
$f\left(2\right)=2+f\left(1\right)=2+1$ $f\left(3\right)=3+f\left(2\right)=3+2+1$
$f\left(4\right)=4+f\left(3\right)=4+3+2+1$
$\therefore f\left(100\right)=100+99+98+\cdots +3+2+1=5050$