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

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.

mxb67

step-by-step explanation: