Suppose a_{1}, a_{2}, ldots, a_{n}, ldots is | vedclass

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(B) Let the first term be $a$ and the common difference be $d$. The sum of the first $n$ terms is $S_{n} = \frac{n}{2}(2a + (n-1)d)$.Given $\frac{S_{5

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$380$

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(B) Let the first term be $a$ and the common difference be $d$. The sum of the first $n$ terms is $S_{n} = \frac{n}{2}(2a + (n-1)d)$.Given $\frac{S_{5}}{S_{9}} = \frac{5}{17}$,we have $\frac{\frac{5}{2}(2a + 4d)}{\frac{9}{2}(2a + 8d)} = \frac{5}{17}$.Simplifying,$\frac{5(a + 2d)}{9(a + 4d)} = \frac{5}{17}$ $\Rightarrow 17(a + 2d) = 9(a + 4d)$ $\Rightarrow 17a + 34d = 9a + 36d$ $\Rightarrow 8a = 2d$ $\Rightarrow d = 4a$.The $15^{th}$ term is $a_{15} = a + 14d = a + 14(4a) = a + 56a = 57a$.Given $110 Since $a$ is a natural number,$a = 2$.Then $d = 4(2) = 8$.The sum of the first ten terms is $S_{10} = \frac{10}{2}(2a + 9d) = 5(2(2) + 9(8)) = 5(4 + 72) = 5(76) = 380$.

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 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 -

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