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.

3 Comment(s) on this Question

Khushi Narang 2020-07-27 09:59:05

Its very helpful me ðthanks

Jaas 2019-05-22 08:32:08

Thanks a lot. It's very helpful for students. ð

Ashutosh 2018-11-05 19:16:44

It's very useful for me Thanks