# C Programming for sum of the series 0.6, 0.06, 0.006, 0.0006, …to n terms

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The given series 0.6,0 .o6,.... is a geometric progression where each element is the previous element divided by 10. So find the sum of the series we have to apply the sum of GP one formula for r less than 1(r=0.1 in our case).

Sum = 6/10 [1- (1/10)n/(1-1/10)]
Sum = 6/9 [1- (1/10)n]
Sum = 2/3[1- (1/10n)]

## Example

#include <stdio.h>
#include <math.h>
int main() {
int n = 6;
float sum = 2*((1 - 1 / pow(10, n)))/3;
printf("sum = %f", sum);
}

## Output

sum = 0.666666
Updated on 09-Aug-2019 11:41:53