Sequences and Series
PYP_JEE_ADV_2025_P1
Grade None
Question:
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 ___.
Step-by-Step Solution
Key Concept: Exponential functional equation implies geometric progression; GP sum formula
$f(x+y)=f(x)f(y)$ with $f>0$ implies $f(x)=r^x$ for some $r>0$.
Let $a_i = a_1 + (i-1)d$. Then $f(a_i) = r^{a_i}$, so $f(a_1), f(a_2), \ldots$ form a geometric progression with ratio $r^d$. Let $b_i = f(a_i)$, $b_1 = f(a_1)$, common ratio $q = r^d$.
$f(a_{31}) = 64f(a_{25}) \Rightarrow b_1 q^{30} = 64 b_1 q^{24} \Rightarrow q^6 = 64 = 2^6 \Rightarrow q = 2$.
$\sum_{i=1}^{50} b_1 \cdot 2^{i-1} = b_1(2^{50}-1) = 3(2^{25}+1)$.
Hmm, $2^{50}-1 = (2^{25}-1)(2^{25}+1)$. So $b_1 = \dfrac{3(2^{25}+1)}{(2^{25}-1)(2^{25}+1)} = \dfrac{3}{2^{25}-1}$.
$\sum_{i=6}^{30} f(a_i) = b_1 \cdot 2^5 \cdot \dfrac{2^{25}-1}{2-1} = \dfrac{3}{2^{25}-1} \cdot 32 \cdot (2^{25}-1) = 3 \times 32 = 96$.
Correct Answer: 96