Suppose a _ 1 , a _ 2 , , a _ n , be an arithmetic progression of...

Given the ratio of the sum of the first five terms to the sum of the first nine terms is $5: 17$,

So, $\dfrac{{S}_{5}}{{S}_{9}} = \dfrac{5}{17} ⇒ \dfrac{\dfrac{5}{2}\left(2a + 4d\right)}{\dfrac{9}{2}\left(2a + 8d\right)} = \dfrac{5}{17}$

$⇒ 17\left(2a + 4d\right) = 9\left(2a + 8d\right)$

$⇒ 34a + 68d = 18a + 72d$

$⇒ d = 4a$

Now ${a}_{15} = a + 14d = 57a$

Also given $110 < {a}_{15} < 120$

$⇒ 110 < 57a < 120$

$⇒ a = 2 ∴ d = 8$

So, ${S}_{10} = \dfrac{10}{2}\left(2 \times 2 + 9 \times 8\right) = 380$