Question

If $n$ A.M. s are introduced between $3$ and $17$ such that the ratio of the last mean to the first mean is $1$, then the value of $n$ is:

• A. $1$
• B. $8$
• C. $4$
• D. None of these

Answer 1

The correct option is A $1$

Let $n$ AM are introduced in $3$ and $17$, they form an AP $3,A_1,A_2,A_3,A_4,....,A_n,17$

Given $\frac{A_n}{A_1}=1$

Total number of terms in AP $=(n+2)$

$a_n=a+(n-1)d{n}^{th}$ term formula

$17=3+(n+1)d\frac{14}{n+1}=d$

$A_n=3+(n+1-1)d=3+nd=3+\frac{14n}{n+1}=\frac{12n+3}{n+1}{A}_1=3+(2-1)d=3+d=3+\frac{14}{n+1}=\frac{3n+17}{n+1}$

Given, $A_n=A_1$

$\frac{17n+3}{n+1}=\frac{3n+17}{n+1}14n=14n=1$