Let a 1, a 2, a 3,ldots be an A.P. and g 1 = a 1, g 2, g 3,ldots be an increasing G.P. If a 1 = a 2 + g 2 = 1 and a 3 + g 3 = 4 then a 10 + g 5 is equal to:

Mathematics · Sequence and Series

JEE Main 2026 — 2 April, Evening Shift — Question 26

Let $\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3,\ldots$ be an A.P. and $\mathbf{g}_1 = \mathbf{a}_1,\mathbf{g}_2,\mathbf{g}_3,\ldots$ be an increasing G.P. If $\mathbf{a}_1 = \mathbf{a}_2 + \mathbf{g}_2 = 1$ and $\mathbf{a}_3 + \mathbf{g}_3 = 4$ then $\mathbf{a}_{10} + \mathbf{g}_5$ is equal to:

Option A:
81
Option B:
76
Option C:
62
Option D:
55
Correct

Answer: D

Step-by-step solution

A.P. : $1, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots$ G.P. : $1, \mathrm{~g}_{2}, \mathrm{~g}_{3}, \ldots$ $\mathrm{a}_{2}+\mathrm{g}_{2}=1 \Rightarrow 1+\mathrm{d}+\mathrm{r}=1 \Rightarrow \mathrm{~d}+\mathrm{r}=0$ $1+2 \mathrm{~d}+\mathrm{r}^{2}=4 \Rightarrow \mathrm{r}^{2}-2 \mathrm{r}-3=0$ $\Rightarrow \mathrm{r}=3,-1$ (reject) $\therefore \mathrm{d}=-3$ $\therefore a_{10}=1+9 d=1+9(-3)=-26$ $\mathrm{g}_{5}=1 . \mathrm{r}^{4}=3^{4}=81$ $\therefore \mathrm{a}_{10}+\mathrm{g}_{5}=55$

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Geometric Progression

The sum 1^3/1 +1^3 + 2^3/1 + 3 +1^3 + 2^3 + 3^3/1 + 3 + 5 +ldots up to 8 terms, is:

Answer: D

Step-by-step solution

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