Question

There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3 : 1. Find the value of n.

Answer 1

Let the n A.M.s between 3 and 17 be $A_1,A_2,A_3,\dots \dots ,A_n$

Then,

ATQ

$\frac{A_n}{A_1}=\frac{3}{1}\dots \dots \left(i\right)$

We know that

$3,A_1,A_2,A_3,\dots \dots ,A_n,17$ are in A.P. of n + 2 terms.

So, 17 is the (n + 2)th terms.

i.e. 17 = 3 + (n + 2 - 1)d [Using $a_n=a+(n-1)d$]

Or $d=\frac{14}{(n+1)}\dots \dots \left(ii\right)$

$\therefore A_n=3+(n+1-1)d$

$=3+\frac{14n}{n+1}=\frac{17n+3}{n+1}\dots \dots \left(iii\right)$

$A_1=3+d=\frac{3n+17}{n+1}\dots \dots \left(iv\right)$

From (i), (iii) and (iv)

$\frac{A_n}{A_1}=\frac{17n+3}{3n+17}=\frac{3}{1}$

$\therefore n=6$

There are 6 A.M. between 3 and 17.