Explanation

Given:

$$ S_n = \sum_{k=1}^{n} a_k = \alpha n^2 + \beta n $$

Term of the series is obtained by:

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

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

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

Using $a_{10} = 59$:

$$ \alpha (19) + \beta = 59 \quad (1) $$

Now,

$$ a_6 = \alpha (11) + \beta $$

Also,

$$ a_1 = \alpha + \beta $$

Given $a_6 = 7a_1$:

$$ \alpha (11) + \beta = 7(\alpha + \beta) $$

$$ 11\alpha + \beta = 7\alpha + 7\beta $$

$$ 4\alpha = 6\beta $$

$$ 2\alpha = 3\beta \Rightarrow \beta = \frac{2\alpha}{3} $$

Substitute in equation (1):

$$ 19\alpha + \frac{2\alpha}{3} = 59 $$

$$ \frac{59\alpha}{3} = 59 $$

$$ \alpha = 3,\quad \beta = 2 $$

Therefore,

$$ \alpha + \beta = 5 $$

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