Suppose a 1 , a 2 , … , a n , … be an arithmetic progression of natural numbers. If the ratio of…

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

So, $\frac{S_5}{S_9}=\frac{5}{17}\Rightarrow \frac{\frac{5}{2}\left(2a+4d\right)}{\frac{9}{2}\left(2a+8d\right)}=\frac{5}{17}$

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

$\Rightarrow 34a+68d=18a+72d$

$\Rightarrow d=4a$

Now $a_{15}=a+14d=57a$

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

$\Rightarrow 110<57\mathrm{a}<120$

$\Rightarrow a=2 \therefore d=8$

So, $S_{10}=\frac{10}{2}\left(2\times 2+9\times 8\right)=380$