Question

Find the sum of all 11 terms of an AP whose middlemost terms is 30?

Answer 1

Let the difference between consecutive terms be D. If the middle term is 30, then the term before it is 30-D, and the term after it is 30+D. So the sum of these three terms would be (30-D) + 30 + (30+D) = 3*30.

Extending this sum to include all 11 terms centered around 30, we see that any addition of D is canceled by a balanced subtraction, leaving you with 11 copies of 30. So the value of the sum is 11*30 = 330.