Find the sum of all the 11 terms of an AP whose middle most term is 30.

Explanation

Let first term of A.P. be a and common difference is d.
Since, the total number of terms n=11, which is odd
`:. ` Middle term `=(n+1)/(2)`th term = `(11-1)/(2)` th term =6th term
`:. T_(6)=30 " " `(given)
`:. A+(6-1)d=30 rArr a+5d=30 " " ...(1)`
Sum of n term of an A.P. `=(n)/(2)[2a+(n-1)d]`
`rArr S_(11)=(11)/(2)[2a+(11-1)d]=(11)/(2)xx2(a+5d)`
`=11xx30 " "` [from (1)]
`=330`.