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

Vedclass pdf generator app on play store
Vedclass iOS app on app store

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

Here, $a=105$ and $d=5$

Here, $a=105$ and $d=5$

$a+(n-1) d=995$

$\Rightarrow 105+(n-1) 5=995$

$\Rightarrow(n-1) 5=995-105=890$

$\Rightarrow n-1=178$

$\Rightarrow n=179$

$\therefore S_{n}=\frac{179}{2}[2(105)+(179-1)(5)]$

$=\frac{179}{2}[2(105)+(178)(5)]$

$=179[105+(89) 5]$

$=179(105+445)$

$=(179)(550)$

$=98450$

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

Similar Questions

The ratio of the sums of $m$ and $n$ terms of an $A.P.$ is $m^{2}: n^{2} .$ Show that the ratio of $m^{ th }$ and $n^{ th }$ term is $(2 m-1):(2 n-1)$

If the $A.M.$ between $p^{th}$ and $q^{th}$ terms of an $A.P.$ is equal to the $A.M.$ between $r^{th}$ and $s^{th}$ terms of the same $A.P.$, then $p + q$ is equal to

  • [AIEEE 2012]

If $2x,\;x + 8,\;3x + 1$ are in $A.P.$, then the value of $x$ will be

Write the first three terms in each of the following sequences defined by the following:

$a_{n}=\frac{n-3}{4}$

For any three positive real numbers $a,b,c$ ; $9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right)$ then

  • [JEE MAIN 2017]