# 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```
• Created
Oct 11, 2014
• Updated
Dec 25, 2014
• Views
10,002
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