If $n$ arithmetic mean 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 

Here $d=\frac{100-a}{n+1}$

Now, $A_1=a+d$ and, $A_n=100-d$ So, $\Rightarrow \frac{A_1}{A_n}=\frac{1}{7}\Rightarrow \frac{a+d}{100-d}=\frac{1}{7}\Rightarrow 7a+8d=100$

   $\Rightarrow 7a+8\left(\frac{100-a}{n+1}\right)=100$       ……….. (i) 

    $\because a+n=33$                                  ……….. (ii)

Now, by Eq. (i) and (ii)    $7n^2-132n-667=0$

$n=23$ and $n=\frac{-29}{7}$ reject.