Answer
(a)
$\text{formula for the $nth$ term of arithmetic sequence.}$
$$
a_{n}=a+(n-1) d
$$
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(b)
the first five terms are $3,8,13,18,$ and $23 .$
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(c)
First Formulas
$$S_{n}=\frac{n}{2}[2 a+(n-1) d] $$
Second Formulas
$$ S_{n}=n\left[\frac{a+a_{n}}{2}\right]$$
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(d)
$$
S_{20}=\frac{20}{2}[(2 \times 3)+(19 \times 5)]=1010
$$
Work Step by Step
(a)
An arithmetic sequence $a_{n}$ is obtained when we start with a number $a$ and add to it a fixed constant $d$ over and over again.
So a formula for the $nth$ term of an arithmetic sequence.
$$
a_{n}=a+(n-1) d
$$
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(b)
The first term is $a=3,$ and the common difference is [ $d=8-3=5$] .
So the $n$th term is
$$
a_{n}=3+(n-1) 5
$$
which can simplifies to $$a_{n}=-2+5 n $$
So the first five terms are
$$a_{1}=-2+(5 \times 1) = -2 + 5 = 3$$ $$a_{2}=-2+(5 \times 2) = -2 + 10 = 8$$ $$a_{3}=-2+(5 \times 3) = -2 + 15 = 13$$ $$a_{4}=-2+(5 \times 4) = -2 + 20 = 18$$ $$a_{5}=-2+(5 \times 5) = -2 + 25 = 23$$
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(c)
$\text{two different formulas for the sum of the first $n$ terms of an arithmetic}$ $\text{ sequence.}$
First Formulas
$$S_{n}=\frac{n}{2}[2 a+(n-1) d] $$
Second Formulas
$$ S_{n}=n\left[\frac{a+a_{n}}{2}\right]$$
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(d)
$\text{to get sum of first 20 terms of the sequence in part (b).}$
$\text{we use the first formula in part $(c)$, so we get}$
$$
S_{20}=\frac{20}{2}[(2 \times 3)+(19 \times 5)]=1010
$$
