Question

Let \[ \sum_{k=1}^{n} a_k = \alpha n^2 + \beta n. \] If \( a_{10} = 59 \) and \( a_6 = 7a_1 \), then \( \alpha + \beta \) is equal to

• A. 3
• B. 5
• C. 7
• D. 12

Hint:

For $S_n = An^2+Bn$, common difference $d=2A$, first term $a=A+B$.

Answer 1

The Correct Option is B

$S_n$ is quadratic $\implies a_n$ is AP.
$a_n = S_n - S_{n-1} = \alpha(2n-1) + \beta$.
$a_{10} = 19\alpha + \beta = 59$.
$a_6 = 11\alpha + \beta$. $a_1 = \alpha + \beta$.
$11\alpha + \beta = 7(\alpha + \beta) = 7\alpha + 7\beta$.
$4\alpha = 6\beta \implies \beta = \frac{2}{3}\alpha$.
Substitute into $a_{10}$: $19\alpha + 0.66\alpha = 59 \implies \frac{59\alpha}{3} = 59 \implies \alpha=3$.
$\beta = 2$.
$\alpha+\beta = 5$.