Question

**Given:** $S_n = \displaystyle\sum_{k=1}^{n} a_k = \alpha n^2 + \beta n$ **Step 1 — Find the general term:** $$a_n = S_n - S_{n-1} = \left(\alpha n^2 + \beta n\right) - \left(\alpha(n-1)^2 + \beta(n-1)\right) = \alpha(2n-1) + \beta$$ (Verification: $a_1 = S_1 = \alpha + \beta$, consistent with $\alpha(2 \cdot 1 - 1) + \beta = \alpha + \beta$. ✓) **Step 2 — Apply condition $a_{10} = 59$:** $$a_{10} = 19\alpha + \beta = 59 \quad \cdots (1)$$ **Step 3 — Apply condition $a_6 = 7a_1$:** $$a_6 = 11\alpha + \beta, \quad a_1 = \alpha + \beta$$ $$11\alpha + \beta = 7(\alpha + \beta)$$ $$4\alpha = 6\beta \implies 2\alpha = 3\beta \implies \beta = \frac{2\alpha}{3} \quad \cdots (2)$$ **Step 4 — Solve equations (1) and (2):** $$19\alpha + \frac{2\alpha}{3} = 59 \implies \frac{57\alpha + 2\alpha}{3} = 59 \implies \frac{59\alpha}{3} = 59 \implies \alpha = 3$$ $$\beta = \frac{2 \times 3}{3} = 2$$ **Step 5 — Compute $\alpha + \beta$:** $$\alpha + \beta = 3 + 2 = 5$$ **Answer: (C) $5$**