If n arithmetic means are inserted between a | vedclass

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Question

$d =\frac{100- a }{ n +1}$

$A _{1}= a + d$

$A _{ n }=100- d$

$\Rightarrow \frac{ A _{1}}{ A _{ n }}=\frac{1}{7} \Rightarrow \frac{ a + d }{10

Vedclass · Accepted Answer

$23$

Vedclass · Answer

$d =\frac{100- a }{ n +1}$

$A _{1}= a + d$

$A _{ n }=100- d$

$\Rightarrow \frac{ A _{1}}{ A _{ n }}=\frac{1}{7} \Rightarrow \frac{ a + d }{100- d }=\frac{1}{7}$

$\Rightarrow 7 a+8 d=100$

$\Rightarrow 7\, a +8\left(\frac{100- a }{ n +1}\right)=100$........$(1)$

$\because a + n =33$.........(2)

$Now,\,by\, Eq. (1) and (2)$

$7 n^{2}-132 n-667=0$

$n =23$ and $n =\frac{-29}{7}$ $reject.$

If $n$ arithmetic means are inserted between a and $100$ such that the ratio of the first mean to the last mean is $1: 7$ and $a+n=33$, then the value of $n$ is

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