Question

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

• A. 6
• B. 8
• C. 4
• D. None of these

Answer 1

The correct option is A

6

Let $A_1,A_2,A_3,A_4,\dots \dots A_n$ be the n arithmetic means between 3 and 17

Let d be the common difference of the A.P. $3,A_1,A_2,A_3,A_4,\dots \dots A_n$ and 17

Then, we have:

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

Now, $A_1=3+d=3+\frac{14}{n+1}=\frac{3n+17}{n+1}$

And, $A_n=3+nd=3+n\left(\frac{14}{n+1}\right)=\frac{17n+3}{n+1}$

$\therefore \frac{A_n}{A_1}=\frac{3}{1}$

$\Rightarrow \frac{\left(\frac{17n+3}{n+1}\right)=\frac{3}{1}}{\left(\frac{3n+17}{n+1}\right)}$

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

$\Rightarrow 17n+3=9n+51$

$\Rightarrow 8n=48$

$\Rightarrow n=6$