Finding the n-th partial sum of an arithmetic progression

Question

Find the $nth$ partial sum given $a_n = 3n+2$ and $n = 10$.

I know that the formula to find the partial sum of an arithmetic series is $S_n = \frac n2 (a_1+a_n)$. However, I do not know how to find $a_1$. Can someone please tell me how to find $a_1$, or can this problem be solved another way?

So far, I have tried to substitute values and have gotten to $S_n = 5(a_1+3n+2)$. But there are two unknowns and I don`t know where to go from there.

Answer 1

$a_1=3\cdot1+2=5$

$a_2=3\cdot2+2=8$

$a_3=3\cdot3+2=11$

$\vdots$

$a_{10}=3\cdot10+2=32$

This is an arithmetic sequence with $d=3$ and $a_1=5$ the formula is

$$S_n=a_1\cdot n+d\cdot\frac{n(n-1)}{2}$$ this is equivalent with your formula: then $a_1=5$, $a_{10}=32$ and $n=10$ $$S_n=\frac{n(a_1+a_n)}{2}$$ $$S_{10}=\frac{10(a_1+a_{10})}{2}$$

Answer 2

Another way is:

$$S_n=\sum_{i=1}^n a_i=\sum_{i=1}^n (3i+2)=3\sum_{i=1}^n i +2 \sum_{i=1}^n 1 =3\frac{n(n+1)}{2}+2n$$

Which gives the same result.

Hope it helps