Question

In n 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

(a) 6
Let $A_1,A_2,A_3,A_4....A_n$ be the n arithmetic means between 3 and 17.
Let d be the common difference of the A.P. 3, $A_1,A_2,A_3,A_4,....A_n$ 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, $A_n=3+nd=3+n\left(\frac{14}{n+1}\right)=\frac{17n+3}{n+1}$

$$\begin{gathered} \therefore \frac{A_n}{A_1}=\frac{3}{1} \\ \Rightarrow \frac{\left({\displaystyle \frac{17n+3}{n+1}}\right)}{\left({\displaystyle \frac{3n+17}{n+1}}\right)}=\frac{3}{1} \\ \Rightarrow \frac{17n+3}{3n+17}=\frac{3}{1} \\ \Rightarrow 17n+3=9n+51 \\ \Rightarrow 8n=48 \\ \Rightarrow n=6 \end{gathered}$$