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JEE Mains · Maths · STD 11 - 8. sequence and series

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 -

  1. A \(290\)
  2. B \(380\)
  3. C \(460\)
  4. D \(510\)
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Answer & Solution

Correct Answer

(B) \(380\)

Step-by-step Solution

Detailed explanation

\(\frac{ S _{5}}{ S _{9}}=\frac{5}{17} \Rightarrow \frac{\frac{5}{2}(2 a+4 d)}{\frac{9}{2}(2 a+8 d)}=\frac{5}{17}\) \(\Rightarrow d=4\,a\) \(a_{15}=a+14 d=57\,a\) Now, \(110< a _{15}<120\) \(110<57\,a < 120\) \(a =2 \therefore d =8\)…
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