Question

The numbers $3^{2\sin z\theta -1},14,3^{4-2\sin 2\theta }$ form first three terms of an A.P. Show that its fifth term is equal to $53$.

Answer 1

Since the numbers are in A.P.
$\therefore 28=3^{2\sin 2\theta -1}+3^{4-2\sin 2\theta }$
or $28=\frac{9^{\sin 2\theta }}{3}=\frac{81}{9^{\sin 2\theta }}$
or $28=\frac{x}{3}+\frac{81}{x}$ where $x=9^{\sin 2\theta }$
or $x^2-84x+243=0$
or $(x-81)(x-3)=0$
$\therefore x=81$ or $3$.
$\therefore x=9^{\sin 2\theta }=81,3$ or $9^2,9^{1/2}$
$\therefore \sin 2\theta =2$ or $1/2$
since $\sin 2\theta$ cannot be greater than $1$ so we choose $\sin 2\theta =\frac{1}{2}$
Hence the terms in A.P. are
$3^0,14,27$ i.e., $1,14,27$.
$\therefore T_5=a+4d=1+4.13=53$.