 QUESTION

# There are n A.M. 's between 3 and 54, such that the 8th mean : (n − 2)th mean :: 3 : 5. Find the value of n.

Hint: Using the n A.M’s and the given numbers: 3 and 54, form an A.P. with (n + 2) terms.
From this A.P., using the formula: the $i^{th}$ term of an A.P. is given by: ${{a}_{i}}=a+\left( i-1 \right)d$, find the 9th and the $(n – 1)^{th}$ terms which are the 8th and the $(n − 2)^{th}$ means, respectively. Equate their ratio to 3 : 5 and find the value of n which is the final answer.

In this question, we are given that there are n A.M.'s between 3 and 54, such that the :
8th mean : (n − 2)th mean :: 3 : 5.

We need to find the value of n.
Suppose the n A.M’s between 3 and 54 are $A_1$, $A_2$, …, $A_n$.
We know that all the A.M.’s between two numbers and the two numbers itself form an arithmetic progression, A.P.

So, there will be an arithmetic progression, A.P. with (n + 2) terms:
3, $A_1$, $A_2$, …, $A_n$, 54

Considering this A.P., we have the following:
First term, a = 3
In this A.P., the ith mean will be the (i + 1)th term of the A.P.
54 is the (n + 2)th term of this A.P.

We know that, the $i^{th}$ term of an A.P. is given by: ${{a}_{i}}=a+\left( i-1 \right)d$
Using the above formula, we will find the (n + 2)th term of this A.P. and equate it to 54 to get the value of the common difference, d.
$54=3+(n+1)d$
$51=(n+1)d$
$d=\dfrac{51}{\left( n+1 \right)}$

Now, 8th mean = 9th term = 3 + 8d
(n – 2)th mean = (n – 1)th term = 3 + (n – 2)d
We are given that 8th mean : (n − 2)th mean :: 3 : 5.
So, $\dfrac{3}{5}=\dfrac{3+8d}{3+\left( n-2 \right)d}$
$9+3\left( n-2 \right)d=15+40d$
$6=\left( 3n-46 \right)d$

Now, substituting the value of d, we will get the following:
$6=\left( 3n-46 \right)\dfrac{51}{\left( n+1 \right)}$
$6n+6=153n-2346$
$147n=2352$
$n=16$

Hence, the value of n for which 8th mean : (n − 2)th mean :: 3 : 5 is 16.