Question

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

Answer 1

Step 1: Finding the value of $n$

Given,

$a_1,a_2,a_3,...$is A.P with common difference $d=2$.

$b_1,b_2,b_3,...$is G.P with common ratio $r=2$

$a_1=b_1=c$ ​ holds for some positive integer $n$ where $2\left(a_1+a_2+a_3+...+a_n\right)=b_1+b_2+b_3+,...+b_n$

$$\begin{gathered} 2\left(a_1+a_2+a_3+...+a_n\right)=b_1+b_2+b_3+,...+b_n \\ 2\left[\frac{n}{2}\left(2a_1+2(n-1)\right)\right]=\frac{b_1\left(2^n-1\right)}{2-1}\left[\because S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)\text{for}\text{A.P}\text{and}{S}_n=\frac{b_1\left(2^n-1\right)}{2-1}\text{for G.P}\right] \\ 2n{a}_1+2n^2-2n=2^n{a}_1-a_1 \\ \left(2n{a}_1-2^n{a}_1+a_1\right)=-2n^2+2n \\ {a}_1\left(2n-2^n+1\right)=2n-2n^2 \\ {a}_1=\frac{2n-2n^2}{\left(2n-2^n+1\right)} \end{gathered}$$

Since $a_1=b_1=c$ and $c\ge 1$

Step 2: Finding $c$

Since $n$ is positive integer, for $$\begin{gathered} n\le 6 \\ \text{i.e}n=\{1,2,3,4,5,6\} \end{gathered}$$ inequality holds

$n=1\Rightarrow c=0$ Rejected $c\ge 1$

$n=2\Rightarrow c<0$ Rejected $c\ge 1$

$n=3\Rightarrow c=\frac{6-18}{6-8+1}=12$ correct

$n=4\Rightarrow c=\text{not an integer}$ Rejected

$n=5\Rightarrow c=\text{not an integer}$ Rejected

$n=6\Rightarrow c=\text{not an integer}$ Rejected

Therefore,$c=12$ for $n=3$

Hence, the number of $c$ holds for some positive integer $n$ where $2\left(a_1+a_2+a_3+...+a_n\right)=b_1+b_2+b_3+,...+b_n$ is one.