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JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Mathematics
Chapter
Sequences and Series

JEE Advanced 2025 · Paper 1 · Mathematics · Sequences and Series

Let ℝ denote the set of all real numbers. Let f: ℝ arrow ℝ be a...

Let R\mathbb{R} denote the set of all real numbers. Let f:R→Rbeafunctionsuchthatf: \mathbb{R} \rightarrow \mathbb{R} be a function such thatf(x)>0forallfor allx \in \mathbb{R},and, andf(x+y)=f(x) f(y)forallfor allx, y \in \mathbb{R}.Lettherealnumbers. Let the real numbers\mathrm{a}_1, \mathrm{a}_2 \ldots, \mathrm{a}_{50}beinanarithmeticprogression.Ifbe in an arithmetic progression. Iff\left(\mathrm{a}_{31}\right)=64 f\left(\mathrm{a}_{25}\right),and, and\sum_{i=1}^{50} f\left(a_i\right)=3\left(2^{25}+1\right)thenthevalueofthen the value of\sum_{i=6}^{30} f(ai)\left(a_i\right) is ________

Correct answer

0

Step-by-step solution

∵f(x+y)=f(x)⋅f(y)⇒f(x)=kx(f(x)>0∀x∈R)∵f(a31)=64f(a25)\begin{aligned} & \because f(x+y)=f(x) \cdot f(y) \\ & \Rightarrow f(x)=k^x \quad(f(x)>0 \forall x \in R) \\ & \because f\left(a_{31}\right)=64 f\left(a_{25}\right)\end{aligned}⇒k(a+30d)=64⋅k(a+24d)⇒k6d=64⇒kd=2\begin{aligned} & \Rightarrow \mathrm{k}^{(\mathrm{a}+30 \mathrm{ d})}=64 \cdot \mathrm{k}^{(\mathrm{a}+24 \mathrm{ d})} \\ & \Rightarrow \mathrm{k}^{6 \mathrm{ d}}=64 \\ & \Rightarrow \mathrm{k}^{\mathrm{d}}=2\end{aligned} \sum_{\mathrm{i}=1}^{50} \mathrm{f}\left(\mathrm{a}_{\mathrm{i}}\right)=\mathrm{f}\left(\mathrm{a}_1\right)+\mathrm{f}\left(\mathrm{a}_2\right)+\ldots \ldots+\mathrm{f}\left(\mathrm{a}_{50}\right) =ka+ka+d+…+ka+49d=ka(k50d−1)kd−1=ka(250−1)=3(225+1)( Given )⇒ka=3225−1\begin{aligned} & =\mathrm{k}^{\mathrm{a}}+\mathrm{k}^{\mathrm{a}+\mathrm{d}}+\ldots+\mathrm{k}^{\mathrm{a}+49 \mathrm{ d}}=\frac{\mathrm{k}^{\mathrm{a}}\left(\mathrm{k}^{50 \mathrm{ d}}-1\right)}{\mathrm{k}^{\mathrm{d}}-1} \\ & =\mathrm{k}^{\mathrm{a}}\left(2^{50}-1\right)=3\left(2^{25}+1\right)(\text{ Given }) \\ & \Rightarrow \mathrm{k}^{\mathrm{a}}=\frac{3}{2^{25}-1}\end{aligned} ∴∑i=630f(ai)=ka+5d+ka+6d+…+ka+29d=ka+5d(k25d−1)kd−1=ka⋅(kd)5(225−1)=3225−1⋅25(225−1)=96\begin{aligned} & \therefore \sum_{\mathrm{i}=6}^{30} \mathrm{f}\left(\mathrm{a}_{\mathrm{i}}\right)=\mathrm{k}^{\mathrm{a}+5 \mathrm{ d}}+\mathrm{k}^{\mathrm{a}+6 \mathrm{ d}}+\ldots+\mathrm{k}^{\mathrm{a}+29 \mathrm{ d}} \\ & \quad=\mathrm{k}^{\mathrm{a}+5 \mathrm{ d}} \frac{\left(\mathrm{k}^{25 \mathrm{ d}}-1\right)}{\mathrm{k}^{\mathrm{d}}-1}=\mathrm{k}^{\mathrm{a}} \cdot\left(\mathrm{k}^{\mathrm{d}}\right)^5\left(2^{25}-1\right) \\ & \quad=\frac{3}{2^{25}-1} \cdot 2^5\left(2^{25}-1\right)=96\end{aligned}$
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