Let $a_{1}, a_{2}, a_{3}, \ldots$ be a sequence of positive integers in arithmetic progression with

Question

Let $a_{1}, a_{2}, a_{3}, \ldots$ be a sequence of positive integers in arithmetic progression with common difference 2. Also, let $b_{1}, b_{2}, b_{3}, \ldots$ 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 $2\left(a_{1}+a_{2}+\ldots+a_{n}\right)=b_{1}+b_{2}+\ldots+b_{n}$ holds for some positive integer $n$, is

Professor PC Thomas Classes · Answer

(1) : $2 \sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}$\ $\Rightarrow 2 \cdot \frac{n}{2}(2 c+(n-1) \cdot 2)=c\left(\frac{2^{n}-1}{2-1}ight)$\ $\Rightarrow \quad n(2 c+2 n-2)=c\left(2^{n}-1ight)$\ $\Rightarrow \quad c\left(2^{n}-2 n-1ight)=n(2 n-2)$ $$\begin{equation*}\Rightarrow \quad c=\frac{2 n(n-1)}{2^{n}-2 n-1} \geq 1 	ag{i} \end{equation*}$$ [As, $c$ must be positive integer and $n \geq 2$ as for $n=1, c=0$, not possible]\ $\Rightarrow \quad 2 n(n-1) \geq 2^{n}-2 n-1$\ $\Rightarrow \quad 2^{n} \leq 2 n^{2}-2 n+2 n+1 \Rightarrow 2^{n} \leq 2 n^{2}+1$\ 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 $n=3, c=\frac{2 \cdot 3 \cdot 2}{8-6-1}=12$\ For $n=4, c=\frac{2 \cdot 4 \cdot 3}{16-8-1}=\frac{24}{7}$,\ not an integer\ For $n=5, c=\frac{2 \cdot 5 \cdot 4}{32-10-1}=\frac{40}{21}$, 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 .\