JEE Advanced 2025 Paper 1Question 1
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 $$\sum_{i=1}^{50}f(a_i)=3(2^{25}+1),$$ then the value of $$\sum_{i=6}^{30}f(a_i)$$ is \_\_\_\_\_\_\_\_.
Functions
Answer$96$
Solution
$f(x)=e^{kx}$, so $f(a_i)$ is a GP with ratio $r=f(d)>0$; $r^6=64\Rightarrow r=2$. $\sum_{1}^{50}=f(a_1)(2^{50}-1)=3(2^{25}+1)\Rightarrow f(a_1)=\frac{3}{2^{25}-1}$. Then $\sum_{6}^{30}=f(a_1)2^5\frac{2^{25}-1}{1}=96$.