Question

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

• A. 21
• B. 22
• C. 23
• D. 24

Answer 1

Answer: (C)

$d=\frac{100-a}{n+1}$
$A_1=a+d$
$A_n=100-d$
$\Rightarrow \frac{A_1}{A_n}=\frac{1}{7}$
$\Rightarrow \frac{a+d}{100-d}=\frac{1}{7}$
$\Rightarrow 7a+8d=100$
$\Rightarrow 7a+8\left(\frac{100-a}{n+1}\right)=100$.....(1)
$\because a+n=33$.........(2)
Now, by Eq. (1) and (2)
$7n^2-132n-667=0$
$n=23$ and $n=\frac{-29}{7}$ reject.