Aspire Faculty ID #18030 · Topic: JEE Main 2026 (23 January Evening Shift) · Just now
JEE Main 2026 (23 January Evening Shift)

Let $\sum a_k = \alpha n^2 + \beta n$. If $a_{10} = 59$ and $a_6 = 7a_4$, then $\alpha + \beta$ is equal to:

Solution

$a_n = S_n - S_{n-1}$

$a_n = (\alpha n^2 + \beta n) - (\alpha(n-1)^2 + \beta(n-1))$

$\Rightarrow a_n = 2\alpha n - \alpha + \beta$

$a_{10} = 59 \Rightarrow 20\alpha - \alpha + \beta = 59$

$a_6 = 7a_4$

Solving ⇒ $\alpha = 3,; \beta = 2$

$\Rightarrow \alpha + \beta = 5$

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