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JEE Advanced Mathematics Sequences and Series 2025 JEE Advanced 2025 (Paper 1)

JEE Advanced Mathematics Question (2025) — Solution

Question

Let R denote the set of all real numbers. Let f: R R be a function such that f(x)>0 for all x R , and f(x+y)=f(x) f(y) for all x, y R . Let the real numbers a _1, a _2 , a _ 50 be in an arithmetic progression. If f ( a _ 31 )=64 f ( a _ 25 ), and _ i=1 ^ 50 f (a_i )=3 (2^ 25 +1 ) then the value of _ i=6 ^ 30 f (a_i ) is ________

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

aligned & f(x+y)=f(x) f(y) \\ & f(x)=k^x (f(x)>0 x R) \\ & f (a_ 31 )=64 f (a_ 25 ) aligned aligned & k ^ ( a +30 ~d ) =64 k ^ ( a +24 ~d ) \\ & k ^ 6 ~d =64 \\ & k ^ d =2 aligned _ i =1 ^ 50 f ( a _ i )= f ( a _1 )+ f ( a _2 )+ + f ( a _ 50 ) aligned & = k ^ a + k ^ a + d + + k ^ a +49 ~d = k ^ a ( k ^ 50 ~d -1 ) k ^ d -1 \\ & = k ^ a (2^ 50 -1 )=3 (2^ 25 +1 )( Given ) \\ & k ^ a = 3 2^ 25 -1 aligned aligned & _ i =6 ^ 30 f ( a _ i )= k ^ a +5 ~d + k ^ a +6 ~d + + k ^ a +29 ~d \\ & = k ^ a +5 ~d ( k ^ 25 ~d -1 ) k ^ d -1 = k ^ a ( k ^ d )^5 (2^ 25 -1 ) \\ & = 3 2^ 25 -1 2^5 (2^ 25 -1 )=96 aligned

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Related: Mathematics — Sequences and Series · All PYQ Banks