Question

Find a1 for the arithmetic series with s20 = 80 And d = 2

Answer 1

we know that
the formula of the arithmetic series  is
Sn=n[2(a1) + (n - 1)d]/2 
where
Sn=80
n=20
d=2
a1=?
then
80=20*[2*a1+19*2]/2----------------> 160=[40*a1+760]
160=[40*a1+760]----------> a1=(160-760)/40------------> a1=-15

the answer is
a1=-15

Answer 2

To solve this, we are going to use the standard formula for arithmetic series $S _{n} = \frac{n}{2} [2a_{1} +(n-1)d]$
where:
$S_{n}$ is the sum of the arithmetic sequence 
$n$ is the number of terms 
$d$ is the difference 
$a_{1}$ is the first term
From our problem we know that $S_{n} =80$, $n=20$, and $d=2$. Lets replace those values in our formula to find $a_{1}$:
$80= \frac{20}{2} [2a_{1} +(20-1)2]$
$80=10(2a_{1} +38)$
$80=20a_{1} +380$
$20a_{1} =-300$
$a_{1} = \frac{-300}{20}$
$a_{1} =-15$

We can conclude that the first term $a_{1}$ of the arithmetic series is $-15$