Question

$\text{List I}$ has four entries and $\text{List II}$ has five entries. Each entry of $\text{List I}$ is to be matched with one or more than one entries of $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}a,b,c,d\text{are in A.P. such that}a+b+c+d=32\text{and}& \text{(P)}& a+b=8\\ & 15ad=7bc\text{, then}(a<b<c<d)& & \\ \text{(B)}& \text{If the first three terms of the sequence}\frac{1}{16},a,b,\frac{1}{6}\text{are in G.P.}& \text{(Q)}& a-b=12\\ & \text{and the last three terms are in H.P., then}(a>0,b>0)& & \\ \text{(C)}& \text{If}{S}_n=(\sqrt{3}+1{)}^{2n}+(\sqrt{3}-1{)}^{2n},n\in N\text{and}& \text{(R)}& a+b=4\\ & S_{n+1}=a{S}_n+b{S}_{n-1}\text{, then}& & \\ \text{(D)}& \text{If}{}^a{C}_b=84,{}^a{C}_{b-1}=36\text{and}{}^a{C}_{b+1}=126\text{, then}& \text{(S)}& a+b=12\\ & & \text{(T)}& a+d=16\end{array}$

Which of the following is the only CORRECT combination?

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

Answer 1

The correct option is B $\text{(A)}\to \text{(P)},\text{(T)}$

$\text{(A)}$
$a,b,c,d$ are in A.P.
Let $a=x-3y,b=x-y,c=x+y,d=x+3y$
Given $a+b+c+d=32$
$\Rightarrow 4x=32$
$\Rightarrow x=8$
Also, $15ad=7bc$
$\Rightarrow 15(x^2-9y^2)=7(x^2-y^2)$
$\Rightarrow 8x^2=128y^2$
$\Rightarrow y^2=4\Rightarrow y=\pm 2$
$\therefore a=2,b=6,c=10,d=14(\because a<b<c<d)$

$\text{(B)}$
$\frac{1}{16},a,b$ are in G.P.
$\Rightarrow 16a^2=b\cdots \left(1\right)$
$a,b,\frac{1}{6}$ are in H.P.
$\Rightarrow \frac{2a}{6a+1}=b\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$
$16a^2=\frac{2a}{6a+1}$
$\Rightarrow 48a^2+8a-1=0$
$\Rightarrow a=-\frac{1}{4},\frac{1}{12}$
$\therefore a=\frac{1}{12}$ and $b=\frac{1}{9}(\because a>0,b>0)$