Question

Suppose $a_1, a_2, \ldots ., a_{ n }, \ldots$ 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}(2 a +4 d )}{\frac{9}{2}(2 a +8 d )}=\frac{5}{17} $
$\Rightarrow d =4 a$
$ a _{15}= a +14 d =57 a$
Now $, 110< a _{15}<120 $
$ \Rightarrow 110<57 a <120$
$ \Rightarrow a =2 \therefore d =8 $
$ S _{10}=\frac{10}{2}(2 \times 2+9 \times 8)=380$