What is the sum of a 52-term arithmetic sequence where the first term is 6 and the last term is 363?

Question

kirbykirby · Answer

So, if you have an arithmetic sequence, it has the form:
$a, a+d, a+2d, a+3d, \ldots, a+nd$ 
Where $a$ is the first term, and $d$ is the common difference. Here, there are $n+1$ terms in the sequence (not $n$ terms), since the first term does not have a "$d$" in it. 

Here, $a=6$. Now, you can write the last term (the 52-nd term) written generally as $a+nd$, more specifically as $6+51d$ since you need 52 terms in total. 
The last term is 363, so $363=6+51d\implies d=7$.

Now, adding up the terms of your sequence:
$6 + (6+7) + (6+2\cdot 7) + \ldots +6+51\cdot 7$


To sum these 52 terms in sigma (summation) notation, you essentially do:
$$\large \sum_{d=0}^{52-1}(6+d\cdot 7)$$

For this, you need the formula:
$$\large \sum_{d=0}^nd=\frac{n(n+1)}{2}$$

Let me know if you need more help.