Sequences and Series – Arithmetic Progression – JEE Main 23 Jan 2026 Shift 2

Solution:

Given sum of n terms $S_n = \alpha n^2 + \beta n$.
The $n^{th}$ term $a_n$ is given by $S_n – S_{n-1}$:
$$ a_n = (\alpha n^2 + \beta n) – (\alpha(n-1)^2 + \beta(n-1)) $$
$$ a_n = \alpha(n^2 – (n-1)^2) + \beta(n – (n-1)) $$
$$ a_n = \alpha(2n – 1) + \beta $$

Given $a_{10} = 59$:
$$ \alpha(2(10) – 1) + \beta = 59 \Rightarrow 19\alpha + \beta = 59 \quad \dots(1) $$

Given $a_6 = 7a_1$:
$$ a_6 = \alpha(11) + \beta = 11\alpha + \beta $$
$$ a_1 = \alpha(1) + \beta = \alpha + \beta $$
$$ 11\alpha + \beta = 7(\alpha + \beta) $$
$$ 11\alpha + \beta = 7\alpha + 7\beta $$
$$ 4\alpha = 6\beta \Rightarrow 2\alpha = 3\beta \Rightarrow \beta = \frac{2}{3}\alpha $$

Substitute $\beta$ into equation (1):
$$ 19\alpha + \frac{2}{3}\alpha = 59 $$
$$ \frac{57\alpha + 2\alpha}{3} = 59 $$
$$ \frac{59\alpha}{3} = 59 \Rightarrow \alpha = 3 $$

Find $\beta$:
$$ \beta = \frac{2}{3}(3) = 2 $$

Value of $\alpha + \beta$:
$$ \alpha + \beta = 3 + 2 = 5 $$

Ans. (3)