Suppose a 1 , a 2 , … , a n , … be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to the sum of first nine terms of the progression is 5 : 17 and 110

Question

Suppose a 1 , a 2 , … , a n , … be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to 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

TestHub · Accepted Answer

Given the ratio of the sum of the first five terms to the sum of the first nine terms is $5 : 17$ ,
So, $\frac{S_{5}}{S_{9}} = \frac{5}{17} \Rightarrow \frac{\frac{5}{2} (2 a + 4 d)}{\frac{9}{2} (2 a + 8 d)} = \frac{5}{17}$
$\Rightarrow 17 (2 a + 4 d) = 9 (2 a + 8 d)$
$\Rightarrow 34 a + 68 d = 18 a + 72 d$
$\Rightarrow d = 4 a$
Now $a_{15} = a + 14 d = 57 a$
Also given $110 < a_{15} < 120$
$\Rightarrow 110 < 57 a < 120$
$\Rightarrow a = 2 ∴ d = 8$
So, $S_{10} = \frac{10}{2} (2 \times 2 + 9 \times 8) = 380$

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