#### i)Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.ii)Find the sum of those integers from 1 to 500 which are multiples of 2 as well as of 5 .iii)Find the sum of those integers from 1 to 500 which are multiples of 2 or 5.

i)

$\\The numbers which are multiples of 2 as well as 5 i.e. multiple of 10.\\ 10, 20, 30 \ldots \ldots . 490.\\ Here first term (a) = 10\\ Common difference (d) = 20 - 10 = 10\\ a\textsubscript{n} = 490\\ a + (n - 1).d = 490\\ 10 + (n - 1)10 = 490\\ (n - 1)10 = 490 - 10\\ \left( n-1 \right)=\frac{480}{10} \\ n = 48 + 1\\ n = 49\\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]=\frac{49}{2}[10+490]=\frac{49}{2}[500] \\ = 49[250] = 12250\\$

ii)

$\\The numbers which are multiples of 2 as well as 5 is the multiple of 10.\\ 10, 20, 30, \ldots .., 500\\ Here first term (a) = 10\\ Common difference (d) = 20 - 10 = 10\\ a\textsubscript{n} = 500 \because a\textsubscript{n}=a + (n - 1)d \\ a + (n - 1)d = 500\\ 10 + (n - 1)10 = 500\\ (n - 1) 10 = 500 - 10\\ n-1=\frac{490}{10} \\ n = 49 + 1\\ n = 50\\ {{S}_{n}}=\frac{n}{2}\left[ a+{{a}_{n}} \right] put n =50\\ 25[510] = 12750\\$

iii)

$\\The numbers which are multiples of 2 are 2, 4, 6, 8, \ldots \ldots , 500\\ Here first term (a\textsubscript{1}) = 2, common difference(d\textsubscript{1}) = 4 - 2, {{n}_{_{1}}}=\frac{500}{2}=250 \\ {{a}_{{{n}_{_{1}}}}} = 500\\ \mathrm{S}_{\mathrm{n}_{1}}=\frac{\mathrm{n}_{1}}{2}\left[\mathrm{a}_{1}+\mathrm{a}_{\mathrm{n}_{1}}\right] \\ \mathrm{S}_{\mathrm{n}_{1}}=\frac{250}{2}[2+500] \\ \mathrm{S}_{\mathrm{n}_{1}}=125[502] \\ \mathrm{S}_{\mathrm{n}_{1}}=62750\\ The numbers which are multiples of 5 are 5, 10, 15, \ldots .., 500\\ Here a\textsubscript{2} = 5, d\textsubscript{2} = 10 - 5 = 5\\ a _{{{n}_{2}}} = 500\\ {{S}_{{{n}_{2}}}}=\frac{{{n}_{2}}}{2}\left[ {{a}_{2}}+{{a}_{{{n}_{2}}}} \right] \Rightarrow {{S}_{{{n}_{2}}}}=\frac{100}{2}\left[ 5+500 \right]=50\left( 505 \right)=25250 \\ The numbers which are multiples of 10 are 10, 20, 30, \ldots .., 50\\ Here a\textsubscript{3} = 10, d\textsubscript{3} = 20 - 10 = 10\\ {{n}_{3}}=\frac{500}{10}=50 \\$

$\\ \mathrm{S}_{\mathrm{n}_{3}}=\frac{\mathrm{n}_{3}}{2}\left[\mathrm{a}_{3}+\mathrm{a}_{\mathrm{n}_{3}}\right] \\\\ \mathrm{S}_{\mathrm{n}_{3}}=\frac{50}{2}[10+500]=25[510]=12750$

The sum of integers from 1 to 500 which are multiples of 2 or 5 = sum of integers which are multiples of 2 + sum of integer which are multiple of 5 - the sum of integer which is multiple of 10

(we subtract the sum of integers multiple of 10 because those integers are multiples of both 2 and 5 )

$=\mathrm{S}_{\mathrm{r}_{1}}+\mathrm{S}_{\mathrm{n}_{2}}-\mathrm{S}_{\mathrm{n}_{3}}=62750+25250-12750$

= 88000 - 12750 = 75250