Question

If $nA{M}^{\prime }s$ are inserted between $1$ and $31$ and ratio of $7^{th}$ and $(n-1{)}^{th}$ $A.M.$ is $5:9$ then $n$ equals ?

• A. $12$
• B. $13$
• C. $14$
• D. $None$

Answer 1

The correct option is C $14$

Since n arithmetic means are inserted between 1 and 31

so Total number of terms=n+2

$\therefore 31=1+(n+2-1)d\to 30=(n+1)d\to d=\frac{30}{n+1}$

AS per question

There Ratio of 7 and (n-1) term is given ,These are nothing but the terms of 8 and n terms according to new AP

$\frac{t_8}{t_n}=\frac{5}{9}\to \frac{1+7d}{1+(n-1)d}=\frac{5}{9}\to 9+63d=5+5(n-1)d\to 9+63\frac{30}{n+1}=5+5(n-1).\frac{30}{n+1}$

$\frac{9n+9+1890}{n+1}=\frac{5n+5+150n-150}{n+1}$

$9n+1899=155n-145$

$\to 146n=2044n=14$