(i) Find the sum of all integers between 100 and 550, which are divisible by 9.
(ii) all integers between 100 and 550 which are not divisible by 9.
(iii) all integers between 1 and 500 which are multiples of 2 as well as of 5.
(iv) all integers from 1 to 500 which are multiplies 2 as well as of 5.
(v) all integers from 1 to 500 which are multiples of 2 or 5.

1-
a=108 as it is the first no between 100 and 550 divisible by 9.

d=9
$a_n$=a+(n-1)d
549=108+(n-1)9
$\frac{441}{9}$=n-1
49+1=n
n=50
Hence no. of terms=50

2-
Sum = sum of all integers between 100 and 550 minus the sum of all integers divisible by 9
sum=

sum=145925-16425=129500
Sum of all integers between 100 and 550 which are not divisible by 9 is 129500
3-
We first find the LCM of 2 and 5 which is 10.
Now all those integers which are multiples of 10 are also the multiples of 2 and 5.
Therefore, multiples of 2 as well as of 5 between 1 and 500 are:
10, 20, 30, ...., 490
Series forms an AP with first term,a=10 and common difference,d=20-10=10.
Let the total number of terms of this AP be n.
Therefore, nth term of AP, $a_n$ = Last term, l= 490

$a_n$ =a(n-1)d= l;
10+(n-1)10=490;
(n-1)10=480;
n-1=48;
n=48+1=49;
n=49

Thus, the sum of n terms of AP is given as:
$S_n$=$\frac{49}{2}$ (10+490);
=$\frac{49}{2}$ (500);
=49×250=12250

4.

a = 10 and l =500 as both the terms are included

$a_n$ =a(n-1)d= l;
10+(n-1)10=500;
(n-1)10=490;
n-1=49;
n=49+1=50;
n=50

Thus, the sum of n terms of AP is given as:
$S_n$=$\frac{50}{2}$ (10+500);
=$\frac{50}{2}$ (510);
=25×510=12750

5.

a = 2 and l =500 as both the terms are included

The series will be 2,4,6,8,10.......500

$a_n$ =a(n-1)d= l;
2+(n-1)2=500;
(n-1)2=498;
n-1=249;
n=249+1=250;
n=250

Thus, the sum of n terms of AP is given as:
$S_n$=$\frac{250}{2}$ (2+500);
=$\frac{250}{2}$ (502);
=125×502=62750