Let \(\mathbb{R}\) denote the set of all real numbers. Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) 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 \(\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{50}\) be in an arithmetic progression. If \(f\left(\mathrm{a}_{31}\right)=64 f\left(\mathrm{a}_{25}\right)\), and \[ \sum_{i=1}^{50} f\left(a_{i}\right)=3\left(2^{25}+1\right), \] then the value of \[ \sum_{i=6}^{30} f\left(a_{i}\right) \] is ____.

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Sequences and Series

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

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

then the value of

i = 6 \sum 30 f (a_{i})

is ____.

Answer: 96

From the functional equation

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

with

f (x) > 0

, we have

f (x) = k^{x}

for some

k > 0

.

Let the AP be

a_{i} = a + (i - 1) d

.

From

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

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

k^{6 d} = 64 = 2^{6}

k^{d} = 2

From

\sum_{i = 1}^{50} f (a_{i}) = 3 (2^{25} + 1)

i = 1 \sum 50 k^{a + (i - 1) d} = k^{a} \cdot \frac{k ^{50 d} - 1}{k ^{d} - 1} = k^{a} \cdot \frac{2 ^{50} - 1}{2 - 1} = k^{a} (2^{50} - 1)

Setting this equal to

3 (2^{25} + 1)

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

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

Now:

i = 6 \sum 30 f (a_{i}) = k^{a + 5 d} \cdot \frac{k ^{25 d} - 1}{k ^{d} - 1} = k^{a} \cdot (k^{d})^{5} \cdot (2^{25} - 1)

= \frac{3}{2 ^{25} - 1} \cdot 2^{5} \cdot (2^{25} - 1) = 3 \cdot 32 = 96

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Numerical answer