Question

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

Answer 1

Answer:

common difference=

d=b-a/n+1

where b=28,a=10,n=11

=28-10/11+1=3/2

AM is even number so,

a+2r×d

=10+2r×3/2

=10+3r, (where r=1)

=13,16,19,22,25

Answer 2

After inserting eleven A.M.s the new series would have a common difference $d=\frac{28-10}{11+1}=\frac{3}{2}$

Hence every even A.M. would be integral as it would be of the form $10+2r\times \frac{3}{2}$ where $r$ is an integer.

$=10+3r$

$=13,16,19,22,25$

Hence, $5$ integral A.M.'s.