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Suppose \mathrm{a_{1}, a_{2}, \ldots, a_{\mathrm{n}}, \ldots} 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 \mathrm{ 110<a_{15}<120,} then the sum of the first ten terms of the progression is equal to

Option: 1

290


Option: 2

380


Option: 3

460


Option: 4

510


Answers (1)

\mathrm{\frac{S_{5}}{S_{g}}=\frac{5}{17} \Rightarrow \frac{\frac{5}{2}(2 a+4 d)}{\frac{9}{2}(2 a+8 d)}=\frac{5}{17}} \\

\mathrm{\Rightarrow d=4 a} \\

\mathrm{a_{15}=a+14 d=57 a} \\

\mathrm{\text { Now } 110<a_{15}<120} \\

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

\mathrm{\Rightarrow a=2 \therefore d=8} \\

\mathrm{S_{10}=\frac{10}{2}(2 \times 2+9 \times 8)=380}

Hence correct option is 2

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Kshitij

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