Question

Suppose $a_1,a_2,\dots .,a_n,\dots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms of 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 -

• A. 290
• B. 380
• C. 460
• D. 510

Answer 1

Answer: (B)

$\frac{S_5}{S_9}=\frac{5}{17}\Rightarrow \frac{\frac{5}{2}(2a+4d)}{\frac{9}{2}(2a+8d)}=\frac{5}{17}$
$\Rightarrow d=4a$
$a_{15}=a+14d=57a$
Now $,110<a_{15}<120$
$\Rightarrow 110<57a<120$
$\Rightarrow a=2\therefore d=8$
$S_{10}=\frac{10}{2}(2\times 2+9\times 8)=380$