Question

Find the sum of first 20 odd natural numbers by A.P. method

Answer 1

Hint: Here the end part of the question is the partial answer only. We know that A.P. stands for arithmetic progression. We know that we have formulas to find the nth term of an A.P. and the sum of n terms of the A.P. we will use those formulas here.
Formulas used:
Sum of first n terms is given by \[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\]
Where,
a = the first term
n = total number of terms
d = common difference
\[{S_n}\]= is the sum

Complete step-by-step answer:
We are given the condition that it should be an A.P.
So now the 20 odd and natural numbers will be 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39.
Now we know that there are a total of 20 terms. The first term is 1 and the common difference is 2.
So we will substitute these values in the formula directly.
\[{S_n} = \dfrac{{20}}{2}\left( {2 \times 1 + \left( {20 - 1} \right)2} \right)\]
We will solve the bracket first,
\[{S_n} = \dfrac{{20}}{2}\left( {2 + 19 \times 2} \right)\]
On multiplying we get,
\[{S_n} = \dfrac{{20}}{2}\left( {2 + 38} \right)\]
We will add the numbers in the bracket,
\[{S_n} = \dfrac{{20}}{2} \times 40\]
On dividing we get,
\[{S_n} = 10 \times 40\]
On multiplying we get,
\[{S_n} = 400\]
This is the sum of the first 20 odd but natural numbers.
So, the correct answer is “400”.

Note: Here note that the words first, odd, natural numbers play a very important role because they decide the numbers of the arithmetic progression. Also note that the numbers are odd and not prime or anything so do select them properly. Also here we know the last term so need not to find the last terms also.