a series with same common difference is known as arithmetic series. the first term of series is 'a' and common difference is d. the series looks like a, a + d, a + 2d, a + 3d, . . . find the sum of series.具有相同共同差异的系列被称为算术系列。系列的第一个术语是“ a ”，共同的区别是d。该系列看起来像a + d，a + 2d，a + 3d...找到系列的总和。
<strong>Input :</strong> a = 1
d = 2
n = 4
<strong>Output :</strong> 16
+ 3 + 5 + 7 = 16
<strong>Input :</strong> a = 2.5
d = 1.5
n = 20
<strong>Output :</strong> 335
Input:
The first line consists of an integer T i.e number of test cases. The first line and only line of each test case consists of three integers a,d,n.
输入：
第一行由整数T即测试用例数组成。每个测试用例的第一行和第一行由三个整数a，d，n组成。
Output:
Print the sum of the series. With two decimal places.
输出：
打印系列的总和。有两位小数。
Constraints:
1
1
约束：
1
1
Example:
Input: 1 2 4 2.5 1.5 20 <strong>Output:</strong> 16.00 335.00
其实，就是一个等差数列的求和问题，不过需要注意的是输出是两位小数。
下面是我的代码实现：
#include
#include
int main()
{
int n,i,j;
scanf("%d",&n);
for(i=0;i<n float a scanf printf return><p>然后竟然出现了错误？一起来看一下：</p><div class="aritcle_card flexRow artxards">
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<p><img src="https://img.php.cn/upload/article/000/000/164/171029420168387.png?x-oss-process=image/resize,p_40" alt="AP系列之和">我试了几次的double类型，也都是这个错误。不是很理解怎么做？这个题目的考察点在哪里哦？</p></n>
