Question

Each entry of $\text{List I}$ is to be matched with one entry of $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& 100(\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{99\cdot 100})\text{equals}& \text{(P)}& 7\\ \text{(B)}& \text{If}x\text{is the arithmetic mean between two real numbers}a\text{and}b\text{,}& \text{(Q)}& 9\\ & y=a^{2/3}\cdot b^{1/3}\text{and}z=a^{1/3}\cdot b^{2/3},\text{then}\frac{y^3+z^3}{xyz}\text{equals}& & \\ \text{(C)}& \text{If}198\text{arithmetic means are inserted between}\frac{1}{4}\text{and}\frac{3}{4}\text{, then}& \text{(R)}& 99\\ & \text{the sum of these arithmetic means is}& & \\ \text{(D)}& \text{If}n\text{is a positive integer such that}n\text{,}\frac{n(n-1)}{2}\text{and}& \text{(S)}& 100\\ & \frac{n(n-1)(n-2)}{6}\text{are in A.P., then the value of}n\text{is}& & \\ & & \text{(T)}& 2\end{array}$

Which of the following is the only CORRECT combination?

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

$\text{(C)}$
Let the inserted A.M.s between $\frac{1}{4}$ and $\frac{3}{4}$ be $A_1,A_2,\dots ,A_{198}$
$\frac{1}{4},A_1,A_2,\dots ,A_{198},\frac{3}{4}$
$A_1+A_2+\cdots +A_{198}=198\left(\frac{\frac{1}{4}+\frac{3}{4}}{2}\right)=99$

$\text{(D)}$
$\frac{2n(n-1)}{2}=n+\frac{n(n-1)(n-2)}{6}$
$\Rightarrow 6(n-1)=6+(n-1)(n-2)$
$\Rightarrow 6n-6=6+n^2-3n+2$
$\Rightarrow n^2-9n+14=0$
$\Rightarrow n=7$ or $2$