Question

There are $n$ arithmetic mean between $3$ and $54$, such that the ratio of the eighth mean and$(n-2)th$ mean is $3:5$. Then the value of $n$ is:

• A. 12
• B. 16
• C. 18
• D. 20

Answer 1

Answer: (B)

16

8th mean = 9 th term in A.P $=t_9=3+8d$

(n-2)th mean = (n-1)th term in A.P $=t_{(n-1)}=3+(n-2)d$

It is given that, $\frac{t_9}{t_{n-1}}=\frac{3}{5}$

$\frac{3+8d}{3+(n-2)d}=\frac{3}{5}$

$9+3nd-6d=15+40d$

$3nd=46d+6$ or $nd=\frac{46}{3}d+2$..........(1)

If there are n arithmetic means between 3 and 54, then (n+2) terms are there in A.P.

hence, $t_{n+2}=3+(n+1)d=54$ or $(n+1)d=nd+d=51$ .......................(2)

By substituting nd from eqn.(1) in eqn(2), n is eliminated
$\frac{46}{3}d+2+d=51\Rightarrow 46d+6+3d=153\Rightarrow 49d=147$
and we get $d=3$.
putting $d=3$ in (2)
$n\left(3\right)+3=51$
$3n=48$

Hence $n=16$