Question

Suppose $a_1,a_2,\cdots a_n,\cdots$ be an arithmetic progression of natural numbers. If the ratio of the sum of first five terms to the sum of first nine terms of the progression is $5:17$ and $110<a_{15}<120$, then the sum of the first ten terms of the progression is equal to

Answer 1

$\therefore a_1,a_2,\cdots a_n\cdots$ be an A.P of natural numbers and
$\frac{S_5}{S_9}=\frac{5}{17}\Rightarrow \frac{\frac{5}{2}[2a_1+4d]}{\frac{9}{2}[2a_1+8d]}=\frac{5}{17}$
$\Rightarrow 34a_1+68d=18a_1+72d$
$\Rightarrow 16a_1=4d$
$\therefore d=4a_1$
And $110<a_{15}<120$
$\therefore 110<a_1+14d<120\Rightarrow 110<57a_1<120$
$\therefore a_1=2(\because a_i\in N)$
$d=8$
$\therefore S_{10}=5[4+9\times 8]=380$