If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean…

Let the A.P. be $a, A_1, A_2...A_n, 100$

Here, common difference, $d=\frac{100-a}{n+1}$

Given $\frac{A_1}{A_n}=\frac{1}{7}\Rightarrow \frac{a+d}{100-d}=\frac{1}{7} ...\left(i\right)$

Also $a+n=33$

From options, when $n=23$, $a=10 \text{and} d=\frac{90}{24}=\frac{15}{4}$

from $\left(i\right)$ $\frac{10+{\displaystyle \frac{15}{4}}}{100-\frac{15}{4}}=\frac{55}{385}=\frac{1}{7}$