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# Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of $ 5 . $

To do:

We have to find the

(i) sum of all

(ii) sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.

(iii) sum of those integers from 1 to 500 which are multiples of 2 or 5.

Solution:

(i) Numbers that are multiples of 2 as well as 5 are the multiples of LCM of 2 and 5.

LCM of 2 and 5 $=2\times5=10$

Numbers divisible by 10 are $10, 20,....., 100, 110,....., 990, 1000,......$

Numbers divisible by 2 and 5 between 1 and 500 are $10, 20, ......,490$

This series is in A.P.

Here,

First term $a=10$

Common difference $d=10$

Last term $a_n=490$

We know that,

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

$490=10+(n-1)10$

$490-10=(n-1)10$

$480=(n-1)10$

$48=n-1$

$n=48+1$

$n=49$

We know that,

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{49}{2}[2 \times 10+(49-1) \times 10]$

$=\frac{49}{2}[20+48 \times 10]$

$=\frac{49}{2}(20+480)$

$=49 \times 250$

$=12250$

The sum of all integers between 1 and 500 which are multiples of 2 as well as 5 is $12250$.

(ii) Numbers that are multiples of 2 as well as 5 are the multiples of LCM of 2 and 5.

LCM of 2 and 5 $=2\times5=10$

Numbers divisible by 10 are $10, 20,....., 100, 110,....., 990, 1000,......$

Numbers divisible by 2 and 5 from 1 to 500 are $10, 20, ......,500$

This series is in A.P.

Here,

First term $a=10$

Common difference $d=10$

Last term $a_n=500$

We know that,

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

$500=10+(n-1)10$

$500-10=(n-1)10$

$490=(n-1)10$

$49=n-1$

$n=49+1$

$n=50$

We know that,

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{50}{2}[2 \times 10+(50-1) \times 10]$

$=25[20+49 \times 10]$

$=25(20+490)$

$=25 \times 510$

$=12750$

The sum of all integers from 1 to 500 which are multiples of 2 as well as 5 is $12750$.

(iii) Sum of integers that are multiples of 2 or 5 $=$ Sum of the multiples of 2 $+$ Sum of multiples of 5 which are not divisible by 2.

Numbers divisible by 2 from 1 to 500 are $2, 4, ......,500$

This series is in A.P.

Here,

First term $a=2$

Common difference $d=4-2=2$

Last term $a_n=500$

We know that,

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

$500=2+(n-1)2$

$500-2=(n-1)2$

$498=(n-1)2$

$249=n-1$

$n=249+1$

$n=250$

We know that,

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{250}{2}[2 \times 2+(250-1) \times 2]$

$=125[4+249 \times 2]$

$=125(4+498)$

$=125 \times 502$

$=62750$

Numbers divisible by 5 from 1 to 500 which are not divisible by 2 are $5, 15, ......,495$

This series is in A.P.

Here,

First term $a=5$

Common difference $d=15-5=10$

Last term $a_n=495$

We know that,

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

$495=5+(n-1)10$

$495-5=(n-1)10$

$490=(n-1)10$

$49=n-1$

$n=49+1$

$n=50$

We know that,

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{50}{2}[2 \times 5+(50-1) \times 10]$

$=25[10+49 \times 10]$

$=25(500)$

$=12500$

The sum of all integers from 1 to 500 which are multiples of 2 or 5 is $75250$.

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