Thank you for reporting, we will resolve it shortly
Q.
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
$ 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.