QJEEAdv20Ma1408: JEE Advanced 2020 Mathematics - Class 11 JEEAdv20Ma1408 Solution: Let a 1 , a 2 , a 3 , .... be a sequence of positive integers…

Given arithmetic progression of positive integers terms a₁, a₂, a₃, ..... having common difference '2' and geometric progression of positive integers
terms b₁, b₂, b₃, .... having common ratio '2' with a₁ = b₁ = c, such that
2(a₁ + a₂ + a₃ + ... + a_n) = b₁ + b₂ + b₃ + ... + b_n

$\Rightarrow 2 \times {n \over 2}[2C + (n - 1)2] = C\left( {{{{2^n} - 1} \over {2 - 1}}} \right)$

$\Rightarrow 2nC + 2{n^2} - 2n = {2^n}.C - C$

$\Rightarrow C[{2^n} - 2n - 1] = 2{n^2} - 2n$

$\because$ $C \in N \Rightarrow 2{n^2} - 2n \ge {2^n} - 2n - 1$

$\Rightarrow 2{n^2} + 1 \ge {2^n} \Rightarrow n \le 6$

and, also C > 0 $\Rightarrow$ n > 2

$\therefore$ The possible values of n are 3, 4, 5, 6

So, at $n = 3,\,C = {{(2 \times 9) - 6} \over {8 - 6 - 1}} = 12$

at, $n = 4,\,C = {{32 - 8} \over {16 - 8 - 1}} = {{24} \over 9} = {8 \over 3} \notin N$

at, $n = 5,\,C = {{50 - 10} \over {32 - 10 - 1}} = {{40} \over {21}} \notin N$

and at, $n = 6,\,C = {{72 - 12} \over {64 - 12 - 1}} = {{60} \over {51}} \notin N$

$\therefore$ The required value of C = 12 for n = 3

so number of possible value of C is 1