Suppose a_1,a_2,,a_n, is an arithmetic progression of natural numbers.

Question

Suppose $a_1,a_2,\ldots,a_n,\ldots$ is an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to the sum of the first nine terms is $5:17$, and $110<a_{15}<120$, then the sum of the first ten terms is equal to:

Accepted Answer

$\dfrac{S_5}{S_9} = \dfrac{5(2a_1+4d)}{9(2a_1+8d)} = \dfrac{5}{17} \Rightarrow 34a_1+68d = 18a_1+72d \Rightarrow 16a_1=4d \Rightarrow d=4a_1.$

$a_{15}=a_1+14d=57a_1$, and $110<57a_1<120 \Rightarrow a_1=2$ (natural number), so $d=8$.

$$S_{10} = \frac{10}{2}\big[2(2)+9(8)\big] = 5(4+72) = 380.$$

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