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

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Question

$\frac{ S _{5}}{ S _{9}}=\frac{5}{17} \Rightarrow \frac{\frac{5}{2}(2 a+4 d)}{\frac{9}{2}(2 a+8 d)}=\frac{5}{17}$

$\Rightarrow d=4\,a$

$a_{15}=a

Vedclass · Accepted Answer

$380$

Vedclass · Answer

$\frac{ S _{5}}{ S _{9}}=\frac{5}{17} \Rightarrow \frac{\frac{5}{2}(2 a+4 d)}{\frac{9}{2}(2 a+8 d)}=\frac{5}{17}$

$\Rightarrow d=4\,a$

$a_{15}=a+14 d=57\,a$

Now, $110< a _{15}<120$

$110<57\,a < 120$

$a =2 	herefore d =8$

$S _{10}=\frac{10}{2}(2 	imes 2+9 	imes 8)=380$

Suppose $a_{1}, a_{2}, \ldots, a_{ n }, \ldots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms of 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 -

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