Question

If $11$ A.M.'s are inserted between $28$ and $10$, then the number of integral A.M.'s is

Answer 1

Given : $11$ A.M.'s are inserted between $28$ and $10$, so
$d=\frac{10-28}{11+1}=-\frac{3}{2}$
Now, let $A_n$ be the $n^{th}$ A.M., so
$A_n=28-\frac{3n}{2}$
For integral A.M.'s, $n$ should be even.
$\therefore n=2,4,6,8,10$

Hence, the number of integral A.M.'s is $5$.