If 11 AM's are inserted between 28 and 10, than find the three middle terms in the series.

Explanation

Let `A_(1),A_(2),A_(3),"….",A_(11)` be 11 AM's between 28 and 10.
If d be the common difference, than
`d=(10-28)/(12)=-(3)/(2)`
Total means `=11(" odd ")`
`therefore` Middle mean `=((11+1)/(2))th =6th =A_(6)`
Then, three middle terms are `A_(5),A_(6),A_(7)`.
`therefore A_(5)=28+5d=28-(15)/(2)=(41)/(2)`
`A_(6)=28+6d=28-9=19`
and `A_(7)=28+7d=28-(21)/(2)=(35)/(2)`