Let ∑_ 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

Correct answer: D. 5 S_n = α n^2 + β n a_n = S_n - S_ n-1 = α(2n-1) + β for n ≥ 1 . a_ 10 = 19α + β = 59 ... (i) a_6 = 11α + β , a_1 = α + β . a_6 = 7a_1 11α + β = 7α + 7β 4α = 6β α = (3β)/(2) ... (ii) Substituting (ii) in (i): 19 · (3β)/(2) + β = 59 (59β)/(2) = 59 β = 2 , α = 3 . α + β = 5 .

JEE Main 202623 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 23 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2026 · 23 January · Evening Shift · Mathematics · Sequences and Series

Let ∑_ k=1 ^ n a_ k =α n^ 2 +β n . If a_ 10 =59 and a_ 6 =7 a_ 1 ,...

Let

\sum_{k=1}^{n} a_{k}=\alpha n^{2}+\beta n

. If

a_{10}=59

and

a_{6}=7 a_{1}

, then

\alpha+\beta

is equal to

Options

Correct answer

D. 5

Step-by-step solution

S_n = \alpha n^2 + \beta n \Rightarrow a_n = S_n - S_{n-1} = \alpha(2n-1) + \beta

for

n \ge 1

a_{10} = 19\alpha + \beta = 59

... (i)

a_6 = 11\alpha + \beta

a_1 = \alpha + \beta

a_6 = 7a_1 \Rightarrow 11\alpha + \beta = 7\alpha + 7\beta \Rightarrow 4\alpha = 6\beta \Rightarrow \alpha = \frac{3\beta}{2}

... (ii)
Substituting (ii) in (i):

19 \cdot \frac{3\beta}{2} + \beta = 59 \Rightarrow \frac{59\beta}{2} = 59 \Rightarrow \beta = 2

\alpha = 3

\alpha + \beta = 5

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