等差数列級数の和を求めるCプログラム

問題

等差数列の合計を求めます。ここで、ユーザーは最初の数、要素の総数、および一般的な差を入力する必要があります。

解決策

等差数列（A.P.）は、連続する2つの数値の差が常に同じである一連の数値です。ここでは、要素の総数はTnとして示されています。

Sum of A.P. Series: Sn = n/2(2a + (n – 1) d)
Tn term of A.P. Series: Tn = a + (n – 1) d

アルゴリズム

等差数列を見つけるには、以下のアルゴリズムを参照してください。

Step 1: Declare variables.
Step 2: Initialize sum=0
Step 3: Enter first number of series at runtime.
Step 4: Enter total number of series at runtime.
Step 5: Enter the common difference at runtime.
Step 6: Compute sum by using the formula given below.
   sum = (num * (2 * a + (num - 1) * diff)) / 2
Step 7: Compute tn by using the formula given below.
   tn = a + (num - 1) * diff
Step 8: For loop
   i = a; i <= tn; i = i + diff
   i. if(i != tn)
      printf("%d + ", i);
   ii. Else,
      printf("%d = %d", i, sum);
Step 9: Print new line

プログラム

以下は、等差数列級数の合計を求めるCプログラムです-

#include <stdio.h>
int main() {
   int a, num, diff, tn, i;
   int sum = 0;
   printf(" enter 1st no of series: ");
   scanf("%d", &a);
   printf(" enter total no's in series: ");
   scanf("%d", &num);
   printf("enter Common Difference: ");
   scanf("%d", &diff);
   sum = (num * (2 * a + (num - 1) * diff)) / 2;
   tn = a + (num - 1) * diff;
   printf("\n sum of A.P series is : ");
   for(i = a; i <= tn; i = i + diff){
      if(i != tn)
         printf("%d + ", i);
      else
         printf("%d = %d", i, sum);
   }
   printf("\n");
   return 0;
}

出力

上記のプログラムを実行すると、次の結果が得られます-

enter 1st no of series: 3
enter total no's in series: 10
enter Common Difference: 5
sum of A.P series is: 3 + 8 + 13 + 18 + 23 + 28 + 33 + 38 + 43 + 48 = 255
enter 1st no of series: 2
enter total no's in series: 15
enter Common Difference: 10
sum of A.P series is: 2 + 12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 + 92 + 102 + 112 + 122 + 132 + 142 = 1080