Alg8.5

How many terms of an A.P $1,4,7,.......\,$ are needed to make the sum 715?

Here $a=1,d=3,S_{n}=715\,$

$S_{n}=\frac{n}{2}[2(1)+(n-1)3]=715\,$

$\frac{n}{2}[3n-1]=715,2n+3n^2-3n=1430,3n^2-n-1430=0\,$

This is a quadratic equation in $n\,$

Solving $n=22,-\frac{65}{3}\,$

The negative number cannot be taken, hence the number of terms $22\,$