JEE Advanced 2025 Paper 1 · Q12 · Sequences & Series

Question

Let $f: \mathbb{R} 	o \mathbb{R}$ satisfy $f(x) > 0$ for all $x$ and $f(x + y) = f(x) f(y)$ for all $x, y \in \mathbb{R}$. Let $a_1, a_2, \ldots, a_{50}$ be in arithmetic progression with $f(a_{31}) = 64\, f(a_{25})$ and $\sum_{i=1}^{50} f(a_i) = 3\,(2^{25} + 1)$. Find $\sum_{i=6}^{30} f(a_i)$.

SolveKar AI · Accepted Answer

From $f(x+y) = f(x) f(y)$ and $f > 0$, $f(x) = c^x$ for some $c > 0$. $f(a_i) = c^{a_i}$ is geometric with common ratio $r = c^d$, where $d = a_2 - a_1$. $f(a_{31}) = 64 f(a_{25})$ $\Rightarrow$ $r^6 = 64$ $\Rightarrow$ $r = 2$ (positive). Sum of all $50$ terms: $f(a_1)(1 + r + \ldots + r^{49}) = f(a_1) \frac{2^{50} - 1}{2-1} = f(a_1)(2^{50} - 1)$. Set equal to $3(2^{25} + 1)$. Note $2^{50} - 1 = (2^{25} - 1)(2^{25} + 1)$. So $f(a_1) = \frac{3(2^{25} + 1)}{(2^{25} - 1)(2^{25} + 1)} = \frac{3}{2^{25} - 1}$. $\sum_{i=6}^{30} f(a_i) = f(a_6)(1 + r + \ldots + r^{24}) = f(a_1) r^5 (2^{25} - 1) = \frac{3}{2^{25} - 1} \cdot 2^5 \cdot (2^{25} - 1) = 3 \cdot 32 = 96$. Answer: $96$.

JEE Advanced 2025 Paper 1 · Q12 · Sequences & Series

Solution