Let a _ 1 , a _ 2 , a _ 3 , . . . . . . . . . be a sequence of...

2\left({a}_{1} + {a}_{2} + . . . . . . . . + {a}_{n}\right) = {b}_{1} + {b}_{2} + . . . . . . . . + {b}_{n}$</p><p>$⇒ 2\dfrac{n}{2}\left[2{a}_{1} + \left(n - 1\right) ⋅ 2\right] = \dfrac{{b}_{1} ⋅ \left({2}^{n} - 1\right)}{2 - 1}$</p><p>$⇒ n\left[2c + 2n - 2\right] = c\left({2}^{n} - 1\right)$</p><p>$⇒ 2n\left[c + n - 1\right] = c\left({2}^{n} - 1\right)$</p><p>$⇒ c\left[{2}^{n} - 2n - 1\right] = 2{n}^{2} - 2n$</p><p>$⇒ c = \dfrac{2{n}^{2} - 2n}{{2}^{n} - 2n - 1} \ge 1 . . . . . . . . . . . . . \left(i\right)$</p><p>$⇒ 2{n}^{2} - 2n \ge {2}^{n} - 2n - 1$</p><p>$⇒ 2{n}^{2} + 1 \ge {2}^{n} ⇒ n \le 6$</p><p>Now put$n = 1, 2, . . . . . . , 6$in equation \left(i\right)$ and using $c \in {I}^{+}$, we get

$c = 12$, when $n = 3$

So, only one value of $c$ is possible.