Let denote the set of all real numbers. Let f : be a function such that f ( x ) 0 for all x , and f ( x + y ) = f ( x ) f ( y ) for all x , y . Let the real numbers a 1 , a 2 , … , a 50 be in an arithmetic progression. If f ( a 31 ) = 64 f (

Question

Let $ℝ$ denote the set of all real numbers. Let $f : ℝ \to ℝ$ be a function such that $f (x) > 0$ for all $x \in ℝ$ , and $f (x + y) = f (x) f (y)$ for all $x, y \in ℝ$ . Let the real numbers $a_{1}, a_{2}, \dots, a_{50}$ be in an arithmetic progression. If $f (a_{31}) = 64 f (a_{25})$ , and $\sum_{i = 1}^{50} f (a_{i}) = 3 (2^{25} + 1),$ then the value of $\sum_{i = 6}^{30} f (a_{i})$ is [2025]

Smart Achievers · Accepted Answer

(96)

Given $f : ℝ \to ℝ$

$\Rightarrow f (x) = k^{x} (f (x) > 0, \forall x \in R)$

$∵$ $f (a_{31}) = 64 f (a_{25})$

$\Rightarrow k^{(a + 30 d)} = 64 \cdot k^{(a + 24 d)} \Rightarrow k^{6 d} = 64 \Rightarrow k^{d} = 2$

$\sum_{i = 1}^{50} f (a_{i}) = f (a_{1}) + f (a_{2}) + \dots + f (a_{50}) = k^{a} + k^{a + d} + \dots + k^{a + 49 d} = \frac{k^{a} (k^{50 d} - 1)}{k^{d} - 1}$ $(∵ k^{d} = 2)$

$= k^{a} (2^{50} - 1) = 3 (2^{25} + 1) (Given)$

$\Rightarrow k^{a} = \frac{3}{2^{25} - 1}$

$∴$ $\sum_{i = 6}^{30} f (a_{i}) = k^{a + 5 d} + k^{a + 6 d} + \dots + k^{a + 29 d}$

$= k^{a + 5 d} (\frac{k^{25 d} - 1}{k^{d} - 1}) = k^{a} . {(k^{d})}^{5} (2^{25} - 1)$

$= \frac{3}{2^{25} - 1} \cdot 2^{5} (2^{25} - 1) = 96$

Solution

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