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

Question

Let $a_{1}, a_{2}, a_{3}, \dots$ be a sequence of positive integers in arithmetic progression with common difference 2. Also, let $b_{1}, b_{2}, b_{3}, \dots$ 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 $2 (a_{1} + a_{2} + \dots + a_{n}) = b_{1} + b_{2} + \dots + b_{n}$ holds for some positive integer $n$ , is ______. [2020]

Smart Achievers · Accepted Answer

(1)

It is given that

$2 (a_{1} + a_{2} + \dots + a_{n}) = b_{1} + b_{2} + \dots + b_{n}$

$\Rightarrow 2 \times \frac{n}{2} (2 c + (n - 2) \cdot 2) = c (\frac{2^{n} - 1}{2 - 1})$ $[∵ a_{1} = c, b_{1} = c]$

$\Rightarrow c (2^{n} - 1 - 2 n) = 2 n^{2} - 2 n$

$\Rightarrow c = \frac{2 n^{2} - 2 n}{2^{n} - 1 - 2 n} \in ℕ$

$So, 2 n^{2} - 2 n \geq 2^{n} - 1 - 2 n$

$\Rightarrow 2 n^{2} + 1 \geq 2^{n} \Rightarrow n < 7$

$∵$ $c \in ℕ \Rightarrow c > 0 \Rightarrow n > 2$

$\Rightarrow n can be 3, 4, 5 or 6$

$Checking c against these values of n$

$When n = 3, c = 14$

$When n = 4, c = \frac{24}{7} which is not possible$

$When n = 5, c = \frac{40}{21} which is not possible$

$When n = 6, c = \frac{60}{51} which is not possible$

$∴$ $we get c = 12 when n = 3$

$Hence, there exists only one value of c which holds the inequality.$

Solution

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