Question

There are n A.M.'s between $3$ and $29$ such that $6$th mean: $(n-1)$the mean:: $3:5$ then find the value of n.

Answer 1

Consider the AMs are $A_1,A_2,\dots ,A_n$. Hence, this creates an AP with $a=3$, $A_{n+1}=29$ and common difference $\frac{29-3}{n+1}=\frac{26}{n+1}$

$\therefore A_k=3+k\frac{26}{n+1}$ where $k\in \{0,n+1\}$

Now,

$A_6:A_{n-1}=\frac{3+6\frac{26}{n+1}}{3+(n-1)\frac{26}{n+1}}$

$\therefore \frac{3}{5}=\frac{3(n+1)+156}{3(n+1)+26(n-1)}$

$\therefore \frac{3}{5}=\frac{3n+159}{29n-23}$

$\therefore 3(29n-23)=5(3n+159)$

$\therefore 87n-69=15n+795$

$\therefore 72n=864$

$\therefore n=12$

Hence, 12 AMs should be inserted.