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First we will need to see the formula to check the sum of first n terms of an A.P. and i.e.,

$ S=\dfrac{n}{2}(a+l).....(1) $

Where S represents sum of first n terms and a represents first term of A.P. and $ l $ represents last term of the A.P.

We are given that first term(a) = 1 and,

Now,

We need to find the last term of the A.P. now and it can be found by using the formula,

$ l=a+(n-1)d $

Where d represents common difference and n represents $ {{n}^{th}} $ term of A.P.

And we know that,

a = 1,

d = 4 – 1 = 3, and

n = 20,

Now applying the mentioned formula for $ l $ , we get

$ \begin{align}

& l=1+(20-1)\times 3 \\

& l=1+(19)\times 3 \\

& l=1+57 \\

& l=58 \\

\end{align} $

Now, we will put this value of $ l $ in equation 1, i.e.

$ \begin{align}

& S=\dfrac{n}{2}(a+l) \\

& S=\dfrac{20}{2}(1+58) \\

& S=10\times 59 \\

& S=590 \\

\end{align} $

Hence we found out that the sum of the first 20 terms of the given A.P. is 590.