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

Answer 1

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

$\text{(C)}$
Given, $S_n=(\sqrt{3}+1{)}^{2n}+(\sqrt{3}-1{)}^{2n}$
$=(4+2\sqrt{3}{)}^n+(4-2\sqrt{3}{)}^n$
$={\alpha }^n+{\beta }^n,$
where $\alpha =4+2\sqrt{3},\beta =4-2\sqrt{3}$
Now, $S_{n+1}={\alpha }^{n+1}+{\beta }^{n+1}$
$=(\alpha +\beta )({\alpha }^n+{\beta }^n)-\alpha \beta ({\alpha }^{n-1}+{\beta }^{n-1})$
$\therefore S_{n+1}=8S_n-4S_{n-1}$
$\Rightarrow a=8,b=-4$

$\text{(D)}$
Given ${}^a{C}_b=84,{}^a{C}_{b-1}=36,{}^a{C}_{b+1}=126$
$\frac{{}^a{C}_b}{{}^a{C}_{b-1}}=\frac{84}{36}$
$\Rightarrow \frac{a-b+1}{b}=\frac{7}{3}$
$\Rightarrow 3a-10b+3=0\cdots \left(1\right)$

$\frac{{}^a{C}_{b+1}}{{}^a{C}_b}=\frac{126}{84}$
$\Rightarrow \frac{a-b}{b+1}=\frac{3}{2}$
$\Rightarrow 2a-5b-3=0\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we get
$a=9,b=3$