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Find the sum of 20 terms of the A.P. 1, 4, 7, 10, …

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Hint: To solve this problem you need to first know that Sum(S) of first n terms of an A.P. is given by the formula, $ S=\dfrac{n}{2}(a+l) $ , where a is the first term of the A.P. and $ l $ is the last term till which we need to find out the sum. To find the last term you can use the formula $ l=a+(n-1)d $ where n represents position of last term in A.P. and d is the common difference of the A.P.

Complete step-by-step answer:
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.

Note: We can also combine the equation $ l=a+(n-1)d $ with $ S=\dfrac{n}{2}(a+l) $ by putting the value of $ l $ in latter equation and we get $ S=\dfrac{n}{2}(2a+(n-1)d) $ and we can now directly find out sum and can skip out the step to find out $ l $ separately. Sometimes students make a mistake while taking values of first term and last term, to find out the sum. The first term is usually the term from which we have to find the sum and last term is the term till which we have to find the sum.