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Q.
If the numbers $3^{2 a - 1}, 14,3^{4 - 2 a}(0 < a < 1)$ are the first three terms of an $AP$, then its fifth term is equal to
Sequences and Series
Solution:
By hypothesis $3^{2 a-1}+3^{4-2 a}=28$. Therefore
$\frac{9^{a}}{3}+\frac{81}{9^{a}}=28$
Substituting $9^{a}=x$, we get $\frac{x}{3}+\frac{81}{x}=28$
$\Rightarrow x^{2}-84 x+243=0$
$ \Rightarrow (x-81)(x-3)=0$
$\Rightarrow x=81$ or $x=3$
This gives $9^{a}=81$ or $9^{a}=3$
$\Rightarrow a=2$ or $\frac{1}{2} $
$ \because 0<\,a<\,1 $
$\Rightarrow a=\frac{1}{2}$
Therefore, the numbers are $1, 14, 27$, which are in AP with common difference $13$ The fifth term
$=1+4 \times 13=53$