Question

Let $a_1,a_2,a_3,\dots$ be a sequence of positive integers in arithmetic progression with common difference $2$. Also, let $b_1,b_2,b_3,\dots$ 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
$2\left(a_1+a_2+\dots +a_n\right)=b_1+b_2+\dots .+b_n$
holds for some positive integer $n$, is

Answer 1

Answer: 1

$\because 2\left(a_1+a_2+a_n\right)=b_1+b_2+\dots +b_n$
$\Rightarrow n[2c+(n-1)2]=c\left(2^n-1\right)$
$\Rightarrow c\left(2^n-2n-1\right)=2n^2-2n$
$\Rightarrow c=\frac{2n^2-2n}{2^n-2n-1}$
$\because c\in N,2n^2-2n\ge 2^n-2n-1$
$\Rightarrow 2n^2+1\ge 2^n$
$\Rightarrow n\le 6$
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 '$c$'.