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Functions

Section: Mathematics

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Question : 4 of 30

Marks: +1, -0

Let

\mathbb{R}

denote the set of all real numbers. Let

f: \mathbb{R} \longrightarrow \mathbb{R}

be a function such that

f(x) > 0

x \in \mathbb{R}

f(x+y) = f(x) f(y)

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}) = 64 f(a_{25})

\sum\limits_{i=1}^{50} f(a_i) = 3 (2^{25}+1)

then the value of

\sum\limits_{i=6}^{30} f(a_i)

[JEE Adv 2025 P1]

Your Answer:

Solution:

f(x) = a^x \; \forall x, y \in R

a_r = a_1 + (r-1) d

. \{Let

d

is common difference of A. P)

\sum\limits_{i=1}^{50} f(a_i) = \sum\limits_{i=1}^{50} a^{\alpha_1+(i-1)d}

= a^{\alpha_1-d} \sum\limits_{i=1}^{50} a^{i d} = a^{\alpha_1-d} = \frac{a^d (1- (a^d)^{50})}{1-a^d}

= a^{\alpha_1-d} \cdot \frac{a^d (1-a^{50d})}{1-a^d}

\Rightarrow a^{\alpha_1} \cdot \frac{1-a^{50d}}{1-a^d} = 3 (2^{25}+1) \;\;\; \cdots (i)

f(a_{31}) = 64 f(a_{25})

\Rightarrow a^{\alpha_1+30d} = 64 \cdot a^{\alpha_1+24d}

\Rightarrow a^{6d} = 64 = 2^6

a^d = 2 \;\;\; \cdots (ii)

Using (i) and (ii)

\Rightarrow \frac{a^{a_1} (1-2^{50})}{1-2} = 3 (2^{25}+1)

\Rightarrow a^{\alpha_1} (2^{25}-1) (2^{25}+1) = 3 (2^{25}+1)

\Rightarrow a^{\alpha_1} = \frac{3}{2^{25}-1}

\text{ Now, } \sum\limits_{i=6}^{30} f(a_i) = a^{\alpha_1-d} \sum\limits_{i=6}^{30} a^{i d} = a^{\alpha_1-d} \cdot \frac{a^{6d} (a^{25d}-1)}{(a^d-1)}

= a^{\alpha_1} \cdot (a^d)^5 \frac{\left| (a^d)^{25} - 1 \right|}{a^d-1}

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

= 96.00

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