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.

Answer 1

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.

Complete step-by-step 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.
This is our final answer.

Note: In this question, it is very important to know that all the A.M.’s between two numbers and the two numbers itself form an arithmetic progression, A.P. Using this we write an arithmetic progression, A.P. with (n + 2) terms. Also note that the ith mean will be the (i + 1)th term of the A.P.