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( a_{1} + a_{2} + + a_{n} ) = b_{1} + b_{2} + + b_{n} holds for some positive integer n, is: Sequence & Series (Mathematics)

$\begin{aligned}& \text { (1) }: 2 \sum_{i=1}^n a_i=\sum_{i=1}^n b_i \\& \Rightarrow 2 \cdot \frac{n}{2}(2 c+(n-1) \cdot 2)=c\left(\frac{2^n-1}{2-1}\right) \\& \Rightarrow n(2 c+2 n-2)=c\left(2^n-1\right) \\& \Rightarrow c\left(2^n-2 n-1\right)=n(2 n-2) \\& \quad \Rightarrow c=\frac{2 n(n-1)}{2^n-2 n-1} \geq 1\end{aligned}$ $\\$ [As, $c$ must be positive integer and $n \geq 2$ as for $n=1, c=0$ , not possible] $\begin{aligned}& \Rightarrow 2 n(n-1) \geq 2^n-2 n-1 \\& \Rightarrow 2^n \leq 2 n^2-2 n+2 n+1 \Rightarrow 2^n \leq 2 n^2+1\end{aligned}$ $\\$ The above inequality holds for $n \leq 6$ . Taking $n=2,3,4,5,6$ using (i), we get

For $n=2, c=\frac{2 \cdot 2 \cdot 1}{4-4-1}=-\mathrm{ve},$ discarded $\\\\$

For $n=3, c=\frac{4-4-3-1}{8-6-1}=12 \\\\$

For $n=4, c=\frac{\frac{8-6-1}{2 \cdot 4 \cdot 3}}{16-9-1}=\frac{24}{7}$ , not an integer $\\\\$

For $n=5, c=\frac{\frac{16 \cdot 5 \cdot 4}{2 \cdot 5 \cdot 10-1}}{32-\frac{40}{21}}$ , not an integer $\\\\$

For $n=6, c=\frac{2 \cdot 6 \cdot 5}{64-12-1}=\frac{60}{51}$ , not an integer $\\$

Thus, $c=12$ is the only possible value. $\\$ Hence, the number of values of $c$ is 1.

Explanation

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