If 3^2^2-1, 14, 3^4-2^2 are the first three terms of an A.P.

Question

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

Accepted Answer

$3^{2\sin^2\alpha-1}+3^{4-2\sin^2\alpha} = 28$. Let $x=3^{2\sin^2\alpha}$: $\dfrac{x}{3}+\dfrac{81}{x}=28 \Rightarrow x^2-84x+243=0 \Rightarrow x=81$ or $3$.

$x=81 \Rightarrow \sin^2\alpha=2$ (impossible); so $x=3$, giving $\alpha=\pi/12$, and $a_1 = 3^{2\sin^2\alpha-1} = 1$, $a_2=14 \Rightarrow d=13$.

$a_6 = 1+5(13) = 66.$

Solutions · 1