Find the sum of all natural numbers lyin | Class 11 Mathematics Chapter Sequence and Series, Sequence and Series NCERT Solutions

Q2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

This sequence forms an A.P.
 
Here, first term, a = 105
 
Common difference, d = 5
 
Here,
 
\begin{align} a + (n - 1)d = 995 \end{align}
 
\begin{align} => 105 + (n - 1)5 = 995 \end{align}
 
\begin{align} =>  (n - 1)5 = 995 - 105 = 890 \end{align}
 
\begin{align} => n -1 = 178 \end{align}
 
\begin{align} => n = 179 \end{align}
 
\begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align}
 
\begin{align} \therefore S_n = \frac {179}{2}\left[2 × (105) + (179 -1)×(5)\right]\end{align}
 
\begin{align} = \frac {179}{2}\left[2(105) + (178)(5)\right]\end{align}
 
\begin{align} = 179\left[105 + (89)5\right]\end{align}
 
\begin{align} = (179)\left[105 + 445\right]\end{align}
 
\begin{align} =179 × 550 \end{align}
 
\begin{align} = 98450 \end{align}
 

Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

 

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What is the correct answer to: Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.?

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.



This sequence forms an A.P.

 

Here, first term, a = 105

 

Common difference, d = 5
...

How do you solve Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5. step by step?

Step-by-step explanation:
• The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995
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Student Discussion

Khushi Narang
Class · · , · Jul 27, 2020
Its very helpful me 👌thanks
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Jaas
Class · · , · May 22, 2019
Thanks a lot. It's very helpful for students. 😊
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