# Class 11 Mathematics Sequence and Series: NCERT Solutions for Question 2

This page focuses on the detailed Sequence and Series question answers for Class 11 Mathematics Sequence and Series, addressing the question: 'Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.'. The solution provides a thorough breakdown of the question, highlighting key concepts and approaches to arrive at the correct answer. This easy-to-understand explanation will help students develop better problem-solving skills, reinforcing their understanding of the chapter and aiding in exam preparation.
Question 2

## 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.