Write a C program to read in two numbers, x and n, and then compute the sum of this geometric progression:
1+x+x2+x3+………….+xn
For example: if n is 3 and x is 5, then the program computes 1+5+25+125.
Print x, n, the sum
Perform error checking. For example, the formula does not make sense for negative exponents – if n is less than 0.
Have your program print an error message if n<0, then go back and read in the next pair of numbers of without computing the sum. Are any values of x also illegal? If so, test for them too.
Algorithm:
Step 1: Start Step 2: read values of x and n, sum  1, i = 1 Step 3: check for n & X i) if n <= 0  x <= 0 ii) print values are not valid iii) read values of x and n Step 4: perform the loop operation i) for(i = 1; i <= n; i++) then follows ii) sum=sum+pow(x, i) Step 5: print sum Step 6: Stop
Program:
#include <stdio.h> #include <conio.h> #include <math.h> void main() { int n, x, i, sum = 0; clrscr(); printf("Enter the limit\n"); scanf("%d", &n); printf("Enter the value of x\n"); scanf("%d", &x); if(x < 0  n < 0) { printf("illegal value"); } else { for(i = 0; i <= n; i++) sum=sum + pow(x, i); } printf("sum=%d", sum); getch(); }
Input & Output:
Enter the limit 4 Enter the value of x sum=31

UpdatedDec 25, 2014

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