sum= a*(1-r^n)/(1-r)
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| sum of first n terms of geometric progression |
Now, we will see a java program to find sum of first n terms of a GP. There is a case in which the above equation fails. When r (common ratio) is 1, the above equation won't work. Then every term in the GP will be same as first term. Therefore, in that case, the sum is a*n. The program is as follows:
import java.util.Scanner;
public class SumGP {
public static void main(String[] arg)
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter first term, common ratio and n (number of terms)");
int a=sc.nextInt(),r=sc.nextInt(),n=sc.nextInt();
System.out.println((r==1)?"Sum is:"+a*n:"Sum is:"+(float)a*(1-Math.pow(r,n))/(1-r));
}
}
