The sum of the fourth and twelfth term of an arithmetic

For any Arithmetic Progression, the n'th term can be written as (a + (n-1)d), where 'a' = first term, 'd' = Common difference between two terms, and 'n' = number of terms.
e.g. 3rd term = a+2d, 1st term = a + 0d = a

The sum of the fourth and twelfth term of an arithmetic progression is 20.
4th Term + 12th Term = 20 ----------> (a+3d) + (a+11d) = 20 ----------> 2a + 14d = 20


What is the sum of the first 15 terms of the arithmetic progression?

Sum of n terms of a arithmetic progression is given by \(Sum = \frac{n(a+l)}{2}\), where 'n' = number of terms, 'a' = first term, and 'l' = last term

\(Sum of first 15 terms = \frac{15(1st term+15th term)}{2}\) ----------> \(\frac{15(a+(a+14d))}{2}\) ------------> \(\frac{15(2a+14d)}{2}\) ---------> \(\frac{(15)(20)}{2}\) -------------> \(150\)