JEE Advanced 2025 · Paper 1 · Mathematics Q12

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MathematicsNumerical+4Question 12 of 48

Let $\mathbb{R}$ denote the set of all real numbers. Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that $f(x) > 0$ for all $x \in \mathbb{R}$ , and $f(x+y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$ .

Let the real numbers $a_1, a_2, \ldots, a_{50}$ be in an arithmetic progression. If $f(a_{31}) = 64f(a_{25})$ , and

$\sum_{i=1}^{50} f(a_i) = 3\left(2^{25} + 1\right),$

then the value of

$\sum_{i=6}^{30} f(a_i)$

is _____.

Your answer

Answer and worked solution

Official answer: 96

Form of f. $f>0$ with $f(x+y)=f(x)f(y)$ , so $f(a_i)=f(a_1)\,r^{i-1}$ where $r=f(d)$ and $d$ is the common difference. The values form a GP.

Ratio. $\frac{f(a_{31})}{f(a_{25})}=r^6=64$ , so $r=2$ .

First term.
$\sum_{1}^{50}f(a_i)=f(a_1)\left(2^{50}-1\right)=3\left(2^{25}+1\right).$
Since $2^{50}-1=(2^{25}-1)(2^{25}+1)$ , this gives $f(a_1)=\frac{3}{2^{25}-1}$ .

Required sum.
$\sum_{6}^{30}f(a_i)=f(a_1)\,2^5\left(2^{25}-1\right)=3\cdot32=96.$

Topics:Geometric progression Types of functions Arithmetic progression

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JEE Advanced 2025 · Paper 1, Mathematics question 12