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$$ \text{Let }\sum_{k=1}^n a_k=\alpha n ^2 +\beta n.$$ If $$a_{10}=59$$ and $$ a_6 = 7a_1,$$ then $$ \alpha+\beta $$ is equal to
$$S_n = \alpha n^2 + \beta n$$.
$$a_n = S_n - S_{n-1} = \alpha(2n-1) + \beta$$.
$$a_{10} = 19\alpha + \beta = 59$$. $$a_1 = \alpha + \beta$$. $$a_6 = 11\alpha + \beta = 7(\alpha + \beta) = 7\alpha + 7\beta$$.
$$4\alpha = 6\beta \implies \alpha = 3\beta/2$$.
From $$19(3\beta/2) + \beta = 59$$
$$59\beta/2 = 59$$, $$\beta = 2$$, $$\alpha = 3$$.
$$\alpha + \beta = 5$$.
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