1. Write the two sequences explicitly

Since $a_1,a_2,a_3,\dots$ is an arithmetic progression of positive integers with common difference $2$ and $a_1=c$, we have $a_k=c+2(k-1).$

Since $b_1,b_2,b_3,\dots$ is a geometric progression of positive integers with common ratio $2$ and $b_1=c$, we have $b_k=c\cdot 2^{k-1}.$

We are given that for some positive integer $n$, $2(a_1+a_2+\cdots+a_n)=b_1+b_2+\cdots+b_n.$

2. Compute the sum of the AP

The sum of first $n$ terms of the AP is $S_A=\frac{n}{2}\bigl(2c+(n-1)\cdot 2\bigr)=\frac{n}{2}(2c+2n-2)=n(c+n-1).$

Therefore, $2S_A=2n(c+n-1).$

3. Compute the sum of the GP

The sum of first $n$ terms of the GP is $S_G=c\frac{2^n-1}{2-1}=c(2^n-1).$

So the condition becomes $2n(c+n-1)=c(2^n-1).$

Rearrange: $2nc+2n(n-1)=c(2^n-1).$

Hence, $c(2^n-1-2n)=2n(n-1).$

So $c=\frac{2n(n-1)}{2^n-1-2n}.$

We need $c$ to be a positive integer.

4. Check small values of $n$

Since denominator is $2^n-1-2n$:

• For $n=1$: $2^1-1-2= -1,$ and directly from original equation, $2a_1=b_1 \implies 2c=c \implies c=0,$ impossible since $c$ is positive.

• For $n=2$: $2^2-1-4=-1,$ giving $c=\frac{2\cdot 2\cdot 1}{-1}=-4,$ impossible.

• For $n=3$: $2^3-1-6=1,$ so $c=\frac{2\cdot 3\cdot 2}{1}=12.$ This is a positive integer, so $c=12$ works.

Let us verify:

• AP: $12,14,16,\dots$ so $a_1+a_2+a_3=12+14+16=42,$ hence $2(a_1+a_2+a_3)=84.$
• GP: $12,24,48,\dots$ so $b_1+b_2+b_3=12+24+48=84.$

Thus $c=12$ indeed satisfies the condition.

5. Show no other $n$ can give an integer $c$

For $n\ge 4$, let $D=2^n-1-2n.$

Then $c=\frac{2n(n-1)}{D}.$

We show $D>2n(n-1)$ for all $n\ge 5$, so then $0<c<1$, impossible for an integer.

Check $n=5$:

so not yet larger. Let us test directly: $c=\frac{40}{21},$ not an integer.

For $n=6$: