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

Welcome to the NCERT Solutions for Class 11th Mathematics - Chapter Sequence and Series. This page offers a step-by-step solution to the specific question from Exercise 2, Question 2: find the sum of all natural numbers lying between....
Question 2

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

Answer

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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