Suppose a_1,a_2, ... ,a_n, ... 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

frac S_5 S_9=(5 / 17) => frac(5 / 2)(2 a+4 d)(9 / 2)(2 a+8 d)=(5 / 17) => d=4 a a_15= a+14 d=57 aNow, 110 => 110 => a=2 Therefore d=8 S_10=(10 / 2)(2 x 2+9 x 8)=380

Suppose a_{1},a_{2},,a_{n}, be an arithmetic progre... | Sequence & Series (Mathematics)

$\frac{\mathrm{S}_5}{\mathrm{~S}_9}=\frac{5}{17} \Rightarrow \frac{\frac{5}{2}(2 \mathrm{a}+4 \mathrm{~d})}{\frac{9}{2}(2 \mathrm{a}+8 \mathrm{~d})}=\frac{5}{17} \\$

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

$\mathrm{a}_{15}=\mathrm{a}+14 \mathrm{~d}=57 \mathrm{a}$

Now, $110<\mathrm{a}_{15}<120\\$

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

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

$\mathrm{~S}_{10}=\frac{10}{2}(2 \times 2+9 \times 8)$ =380

Explanation

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