Let sum k =1 n a k =α n 2 +β n . If a 10 =59 and a 6 =7 a 1 then α+β is equal to

Mathematics · Sequence and Series

JEE Main 2026 — 23 January, Evening Shift — Question 20

Let $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{k}}=\alpha \mathrm{n}^{2}+\beta \mathrm{n}$ . If $\mathrm{a}_{10}=59$ and $\mathrm{a}_{6}=7 \mathrm{a}_{1}$ then $\alpha+\beta$ is equal to

Option A:
$12$
Option B:
$3$
Option C:
$5$
Correct
Option D:
$7$

Answer: C

Step-by-step solution

$\quad \mathrm{a}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}$

$=\left(\alpha \mathrm{n}^{2}+\beta \mathrm{n}\right)-\left(\alpha(\mathrm{n}-1)^{2}+\beta(\mathrm{n}-1)\right)$

$\mathrm{a}_{59} \Rightarrow 19 \alpha+\beta=59$

$a_{6}=7 a_{1} \Rightarrow 11 \alpha+\beta=7(\alpha+\beta)$

$\Rightarrow 2 \alpha=3 \beta$

$\mathrm{a}=3, \beta=2$

$\alpha+\beta=5$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Evening Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression