Question

Let $a_{1}, a_{2}, a_{3}, \ldots$ be a sequence of positive integers in arithmetic progression with common difference $2$. Also, let $b_{1}, b_{2}, b_{3}, \ldots$ be a sequence of positive integers in geometric progression with common ratio $2$ . If $a_{1}=b_{1}=c$, then the number of all possible values of $c$, for which the equality
$2\left(a_{1}+a_{2}+\ldots+a_{n}\right)=b_{1}+b_{2}+\ldots .+b_{n}$
holds for some positive integer $n$, is

Answer 1

$\because 2\left(a_{1}+a_{2}+a_{n}\right)=b_{1}+b_{2}+\ldots+b_{n} $
$\Rightarrow n[2 c+(n-1) 2]=c\left(2^{n}-1\right) $
$\Rightarrow c\left(2^{n}-2 n-1\right)=2 n^{2}-2 n $
$\Rightarrow c=\frac{2 n^{2}-2 n}{2^{n}-2 n-1} $
$\because c \in N, 2 n^{2}-2 n \geq 2^{n}-2 n-1 $
$\Rightarrow 2 n^{2}+1 \geq 2^{n} $
$\Rightarrow n \leq 6 $
also $ c > 0 \Rightarrow n > 2$
So possible values of $n$ are $3,4,5$ and $6$
when $n=3, c=12$
$n=4, c=\frac{24}{7}$ (not possible)
$n=5, c=\frac{40}{21}$ (not possible)
$n=6, c=\frac{60}{51}$ (not possible)
So, there exists only one value of '$c$'.