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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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