JEE PYQ: Sequence And Series Question 30
Question 30 - 2020 (05 Sep Shift 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. for some $\alpha$, then the sixth term of this A.P. is:
(1) 66
(2) 81
(3) 65
(4) 78
Show Answer
Answer: (1)
Solution
Given that $3^{2\sin 2\alpha - 1}$, 14, $3^{4-2\sin 2\alpha}$ are in A.P.
So, $3^{2\sin 2\alpha - 1} + 3^{4-2\sin 2\alpha} = 28$
$\Rightarrow \frac{x}{3} + \frac{81}{x} = 28$ where $x = 3^{2\sin 2\alpha}$
$\Rightarrow x^2 - 84x + 243 = 0 \Rightarrow x = 81, x = 3$
When $x = 81 \Rightarrow \sin 2\alpha = 2$ (Not possible)
When $x = 3 \Rightarrow \alpha = \frac{\pi}{12}$
$\therefore a = 3^0 = 1, d = 14 - 1 = 13$
$a_6 = a + 5d = 1 + 65 = 66$
BETA
