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vlabodo [156]
3 years ago
12

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

Mathematics
2 answers:
AfilCa [17]3 years ago
5 0
We know that
the <span>formula of the arithmetic series  is
</span>Sn=<span>n[2(a1) + (n - 1)d]/2 
where
Sn=80
n=20
d=2
a1=?
then
80=20*[2*a1+19*2]/2----------------> </span>160=[40*a1+760]
160=[40*a1+760]----------> a1=(160-760)/40------------> a1=-15
<span>
the answer is
a1=-15

</span>
Evgesh-ka [11]3 years ago
4 0
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 
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Vedmedyk [2.9K]
We know that
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Group terms that contain the same variable, and move the constant to the opposite side of the equation

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3 years ago
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