$$ \text{Let }\sum_{k=1}^n a_k=\alpha n ^2 +\beta n.$$ 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 $$19(3\beta/2) + \beta = 59$$: $$57\beta/2 + \beta = 59$$, $$59\beta/2 = 59$$, $$\beta = 2$$, $$\alpha = 3$$.

$$\alpha + \beta = 5$$. The answer is Option 3: 5.