Question

If $3^{2\sin 2\alpha -1},14$ and $3^{4-2\sin 2\alpha }$ are the first three terms of an A.P. for some $\alpha$ , then the sixth term of this A.P. is

• A. $65$
• B. $81$
• C. $78$
• D. $66$

Answer 1

The correct option is D

$66$

Explanation for the correct option:

Step 1: If $3^{2\sin 2\alpha -1},14$ and $3^{4-2\sin 2\alpha }$ are the first three terms of an A.P.

$a,b,careinA.Pthen2b=a+c$

Given that, $3^{2\sin 2\alpha -1}+3^{4-2\sin 2\alpha }=28$

$\Rightarrow$ $\frac{9^{\sin 2\alpha }}{3}+\frac{81}{9^{\sin 2\alpha }}=28$

Let $9^{\sin 2\alpha }=t$

Step 2. Put the value of $9^{\sin 2\alpha }=t$ in above expression:

$\frac{t}{3}+\frac{81}{t}=28$

$\Rightarrow$ $t^2-84t+243=0$

$\Rightarrow$ $t^2-81t-3t+243=0$

$\Rightarrow$$t(t-81)-3(t-81)=0$

$\Rightarrow$ $(t-81)(t-3)=0$

$\Rightarrow$ $t=81,3$

Step 3. Put the value of $t$ in $3^{2\sin 2\alpha }$:

$3^{2\sin 2\alpha }=3^1,3^4$

$\Rightarrow$ $2\sin 2\alpha =1,4$

$\Rightarrow$ $\sin 2\alpha =\frac{1}{2},2$

So, First term, $a=3^{2\sin 2\alpha -1}$

$a=1$

Second term $=14$

and Common difference, $d=13$

$\begin{array}{rcl}\therefore T_6& =& a+5d\\ & =& 1+5\times 13\\ & =& 66\end{array}$

Hence, Option ‘$D$’ is Correct.