JEE Main 2026 — Jan 23, Shift 2 — Mathematics solution

JEE Main2026Jan 23, Shift 2Mathematics

Let $k = 1 \sum n a_{k} = α n^{2} + β n$ . If $a_{10} = 59$ and $a_{6} = 7 a_{1}$ , then $α + β$ is equal to

Solution

$a_{n} = S_{n} - S_{n - 1} = (α n^{2} + β n) - (α (n - 1)^{2} + β (n - 1)) = α (2 n - 1) + β$ .

So $a_{n} = 2 α n + (β - α)$ , an arithmetic progression.

$a_{10} = 19 α + β = 59$ ...(1)

$a_{1} = α + β$ , $a_{6} = 11 α + β$ .

$a_{6} = 7 a_{1} \Rightarrow 11 α + β = 7 α + 7 β \Rightarrow 4 α = 6 β \Rightarrow 2 α = 3 β$ ...(2)

From (2): $β = \frac{2 α}{3}$ . Into (1): $19 α + \frac{2 α}{3} = 59 \Rightarrow \frac{59 α}{3} = 59 \Rightarrow α = 3$ , $β = 2$ .

$∴ α + β = 5$ .

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