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Since [tex]a_1,a_2,a_3,\cdots[/tex] are in arithmetic progression,
[tex]a_2 = a_1 + 2[/tex]
[tex]a_3 = a_2 + 2 = a_1 + 2\cdot2[/tex]
[tex]a_4 = a_3+2 = a_1+3\cdot2[/tex]
[tex]\cdots \implies a_n = a_1 + 2(n-1)[/tex]
and since [tex]b_1,b_2,b_3,\cdots[/tex] are in geometric progression,
[tex]b_2 = 2b_1[/tex]
[tex]b_3=2b_2 = 2^2 b_1[/tex]
[tex]b_4=2b_3=2^3b_1[/tex]
[tex]\cdots\implies b_n=2^{n-1}b_1[/tex]
Recall that
[tex]\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n[/tex]
[tex]\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2[/tex]
It follows that
[tex]a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)[/tex]
so the left side is
[tex]2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n[/tex]
Also recall that
[tex]\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}[/tex]
so that the right side is
[tex]b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)[/tex]
Solve for [tex]c[/tex].
[tex]2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}[/tex]
Now, the numerator increases more slowly than the denominator, since
[tex]\dfrac{d}{dn}(2n(n-1)) = 4n - 2[/tex]
[tex]\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2[/tex]
and for [tex]n\ge5[/tex],
[tex]2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2[/tex]
This means we only need to check if the claim is true for any [tex]n\in\{1,2,3,4\}[/tex].
[tex]n=1[/tex] doesn't work, since that makes [tex]c=0[/tex].
If [tex]n=2[/tex], then
[tex]c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0[/tex]
If [tex]n=3[/tex], then
[tex]c = \dfrac{12}{2^3 - 6 - 1} = 12[/tex]
If [tex]n=4[/tex], then
[tex]c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N[/tex]
There is only one value for which the claim is true, [tex]c=12[/tex].