JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Sequences and Series

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 ____

Answer: 1

Solution

We have

a_{k} = c + 2 (k - 1)

and

b_{k} = c \cdot 2^{k - 1}

.

The given equation becomes:

2 \cdot \frac{n}{2} (2 c + (n - 1) \cdot 2) = c (2^{n} - 1)

n (2 c + 2 n - 2) = c (2^{n} - 1)

2 n c + 2 n^{2} - 2 n = c \cdot 2^{n} - c

c (2^{n} - 2 n - 1) = 2 n^{2} - 2 n

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

For

c

to be a positive integer, we need

2 n^{2} - 2 n \geq 2^{n} - 2 n - 1

, which gives

2 n^{2} + 1 \geq 2^{n}

.

This holds for

n \leq 6

. Checking

n = 1, 2, 3, 4, 5, 6

Only

c = 12

works, so the answer is 1.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Numerical answer