The third term of an arithmetic progression is 6. Then the sum of first five…

Given:
Third term of an Arithmetic Progression (AP) = 6
Let the AP be:
a − 2d, a − d, a, a + d, a + 2d
Here, the third term is:
a = 6
So the first five terms become:
6 − 2d, 6 − d, 6, 6 + d, 6 + 2d
Sum of first five terms:
= (6 − 2d) + (6 − d) + 6 + (6 + d) + (6 + 2d)
The d terms cancel out:
= 6 + 6 + 6 + 6 + 6
= 30