The sum of the third and the ninth term of an AP is 10. Find a possible sum of the first 11 terms of this AP?
T3 + T9 = 10
a+2d + a+ 8d = 10
2a+10d=10--------- (1)
Sum to 11 terms = 11/2 (2a+10d)
From 1, sum to 11 terms =11/2 (10) = 55