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

$\text{(A)}$
$100(\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{99\cdot 100})$
$=100[(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdots +(\frac{1}{99}-\frac{1}{100})]$
$=100[1-\frac{1}{100}]$
$=99$

$\text{(B)}$
$\frac{y^3+z^3}{xyz}=\frac{a^{2}b+a{b}^2}{\left(\frac{a+b}{2}\right)(a\times b)}=2$