The sum of third and ninth term of an A.P is 8. Find the sum of the first 11 terms of the progression.

The sum of third and ninth term of an A.P is 8. Find the sum of the first 11 terms of the progression.

Let the first term of AP = a

Common difference = d

Third term  = a+2d

Ninth term = a+8d

Third term + Ninth term  = 8

2a+10d = 8

We know sum of an AP = $$\dfrac{n}{2}\left[2a+\left(n-1\right)d\right]$$

The sum of the first 11 terms = $$\dfrac{11}{2}\left[2a+10d\right]$$

We know that 2a+10d = 8

$$\dfrac{11\times8}{2}=44$$