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# 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 _____

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Solution

## 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$${a}_{1}={b}_{1}=c$ ​ holds for some positive integer $n$ where $2\left({a}_{1}+{a}_{2}+{a}_{3}+...+{a}_{n}\right)={b}_{1}+{b}_{2}+{b}_{3}+,...+{b}_{n}$$2\left({a}_{1}+{a}_{2}+{a}_{3}+...+{a}_{n}\right)={b}_{1}+{b}_{2}+{b}_{3}+,...+{b}_{n}\phantom{\rule{0ex}{0ex}}2\left[\frac{n}{2}\left(2{a}_{1}+2\left(n-1\right)\right)\right]=\frac{{b}_{1}\left({2}^{n}-1\right)}{2-1}\left[\because {S}_{n}=\frac{n}{2}\left(2{a}_{1}+\left(n-1\right)d\right)\text{for}\text{A.P}\text{and}{S}_{n}=\frac{{b}_{1}\left({2}^{n}-1\right)}{2-1}\text{forG.P}\right]\phantom{\rule{0ex}{0ex}}2n{a}_{1}+2{n}^{2}-2n={2}^{n}{a}_{1}-{a}_{1}\phantom{\rule{0ex}{0ex}}\left(2n{a}_{1}-{2}^{n}{a}_{1}+{a}_{1}\right)=-2{n}^{2}+2n\phantom{\rule{0ex}{0ex}}{a}_{1}\left(2n-{2}^{n}+1\right)=2n-2{n}^{2}\phantom{\rule{0ex}{0ex}}{a}_{1}=\frac{2n-2{n}^{2}}{\left(2n-{2}^{n}+1\right)}$Since ${a}_{1}={b}_{1}=c$ and $c\ge 1$Step 2: Finding $c$Since $n$ is positive integer, for $n\le 6\phantom{\rule{0ex}{0ex}}\text{i.e}n=\left\{1,2,3,4,5,6\right\}$ inequality holds$n=1⇒c=0$ Rejected $c\ge 1$$n=2⇒c<0$ Rejected $c\ge 1$$n=3⇒c=\frac{6-18}{6-8+1}=12$ correct$n=4⇒c=\text{notaninteger}$ Rejected $n=5⇒c=\text{notaninteger}$ Rejected $n=6⇒c=\text{notaninteger}$ 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.

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