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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)]

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

sum = 0.666666

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