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(A) The sum of the arithmetic progression is $S_n = \frac{n}{2}(2a_1 + (n-1)d) = \frac{n}{2}(2c + (n-1)2) = n(c + n - 1)$.Given $2(S_n) = b_1 + b_2 +

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(A) The sum of the arithmetic progression is $S_n = \frac{n}{2}(2a_1 + (n-1)d) = \frac{n}{2}(2c + (n-1)2) = n(c + n - 1)$.Given $2(S_n) = b_1 + b_2 + \ldots + b_n$,where $b_n$ is a geometric progression with $b_1 = c$ and $r = 2$.The sum of the geometric progression is $T_n = c(2^n - 1) / (2 - 1) = c(2^n - 1)$.Substituting these into the given equation: $2n(c + n - 1) = c(2^n - 1)$.$2nc + 2n^2 - 2n = c(2^n - 1)$.$2n^2 - 2n = c(2^n - 1 - 2n)$.$c = \frac{2n^2 - 2n}{2^n - 2n - 1}$.For $c$ to be a positive integer,$2n^2 - 2n > 0$ and $2^n - 2n - 1 > 0$.For $n=1$,$c = 0 / (2-2-1) = 0$ (not a positive integer).For $n=2$,$c = (8-4) / (4-4-1) = -4$ (not positive).For $n=3$,$c = (18-6) / (8-6-1) = 12 / 1 = 12$.For $n=4$,$c = (32-8) / (16-8-1) = 24 / 7$ (not an integer).For $n=5$,$c = (50-10) / (32-10-1) = 40 / 21$ (not an integer).For $n=6$,$c = (72-12) / (64-12-1) = 60 / 51$ (not an integer).For $n \geq 7$,$2^n - 2n - 1 > 2n^2 - 2n$,so $c Thus,the only positive integer value for $c$ is $12$,which corresponds to $n=3$.Therefore,the number of possible values of $c$ is $1$.

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