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Question: What is the sum of first 20 odd natural numbers? A) \(100\) B) \(210\) C) \(400\) D) \(420\)...

What is the sum of first 20 odd natural numbers?
A) 100100
B) 210210
C) 400400
D) 420420

Explanation

Solution

In this question, we have to find the sum of first 2020 odd natural numbers. First, find the first 20 odd numbers. To find the 20th20^{th} odd number, use the A.P formula of general term. After you have all the terms, use the formula of sum of A.P to find the sum. Put all the values and simplify. You will get your answer.

Formula used: 1) an=a+(n1)d{a_n} = a + \left( {n - 1} \right)d
2) Sn=n2(2a+(n1)d){S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) or Sn=n2(a+an){S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)

Complete step-by-step solution:
We have to find the sum of the first 2020 odd natural numbers. So, first we will find the first 2020 odd natural numbers. We know that the first few odd numbers are 1,3,5,7,...1,3,5,7,... but we don’t know which is the last number? So, we will find the last and the 20th{20^{th}} odd numbers using the A.P formula of general term.
Finding the 20th{20^{th}} term,
a20=a+(n1)d\Rightarrow {a_{20}} = a + \left( {n - 1} \right)d
Putting a=1,a = 1, n=20,n = 20, d=2d = 2,
a20=1+(201)2\Rightarrow {a_{20}} = 1 + \left( {20 - 1} \right)2
Solving to find the answer,
a20=1+19×2\Rightarrow {a_{20}} = 1 + 19 \times 2
a20=39\Rightarrow {a_{20}} = 39
Hence, 20th{20^{th}} odd number is 3939.
Next, we will find the sum of these first 2020 odd natural numbers. We will use the formula Sn=n2(a+an){S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right) to find the required sum.
Put n=20,n = 20, aa is the first term. So, a=1a = 1 and an{a_n} is the last term. an=39{a_n} = 39.
Putting all the values in the formula –
S20=202(1+39)\Rightarrow {S_{20}} = \dfrac{{20}}{2}\left( {1 + 39} \right)
Simplifying to find the sum,
S20=10×40=400\Rightarrow {S_{20}} = 10 \times 40 = 400

\therefore The sum of first 2020 odd natural numbers is option C)400400.

Note: There is one more formula to find the sum of A.P. Sn=n2(2a+(n1)d) \Rightarrow {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right). Using this formula, you need not even find the 20th{20^{th}} term first. You can directly put all the information that you have about the first term, common difference and the number of terms whose sum is to be found.
Using this formula to find the answer to the above question,
S20=202(2×1+(201)2)\Rightarrow {S_{20}} = \dfrac{{20}}{2}\left( {2 \times 1 + \left( {20 - 1} \right)2} \right)
Simplifying to solve further,
S20=10×(2+38)\Rightarrow {S_{20}} = 10 \times \left( {2 + 38} \right)
S20=10×40=400\Rightarrow {S_{20}} = 10 \times 40 = 400
Therefore, the sum of the first 2020 odd natural numbers is 400400.