# Find the sum of all integers between 100 and 550, which are divisible by 9.

Question:

Find the sum of all integers between 100 and 550, which are divisible by 9.

Solution:

(i) In this problem, we need to find the sum of all the multiples of 9 lying between 100 and 550.

So, we know that the first multiple of 9 after 100 is 108 and the last multiple of 9 before 550 is 549.

Also, all these terms will form an A.P. with the common difference of 9.

So here,

First term (a) = 108

Last term (l) = 549

Common difference (d) = 9

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$549=108+(n-1) 9$

$549=108+9 n-9$

$549=99+9 n$

$549-99=9 n$

Further simplifying,

$450=9 n$

$n=\frac{450}{9}$

$n=50$

Now, using the formula for the sum of n terms,

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

We get,

$S_{n}=\frac{50}{2}[2(108)+(50-1) 9]$

$=25[216+(49) 9]$

$=25(216+441)$

$=25(657)$

$=16425$

Therefore, the sum of all the multiples of 9 lying between 100 and 550 is

(ii)
In this problem, we need to find the sum of all the integers lying between 100 and 550 which are not multiples of 9.
So, we know that the sum of all the multiples of 9 lying between 100 and 550 is 16425.
The sum of all the integers lying between 100 and 550 which are not multiples of 9 is

$[101+102+103+\ldots \ldots+549]-16425$

$=[(1+2+\ldots .+549)-(1+2+\ldots .100)]-16425$

$=\left[\left(\frac{549 \times 550}{2}\right)-\left(\frac{100 \times 101}{2}\right)\right]-16425$

$=150975-5050-16425$

$=129500$

(iii)

Integers between 1 and 500 which are multiples of both 2 and 5 are

10, 20, 30, ....., 490.

$a_{n}=490$

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

$\Rightarrow(n-1) 10=480$

$\Rightarrow n=49$

Now,

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

$S_{49}=\frac{49}{2}[20+(48) 10]$

$S_{49}=\frac{49}{2}[500]$

$S_{49}=49 \times 250=12250$

(iv)

Integers from 1 and 500 which are multiples of both 2 and 5 are

10, 20, 30, ....., 500.

$a_{n}=500$

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

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

$\Rightarrow n=50$

Now,

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

$S_{50}=\frac{50}{2}[20+(49) 10]$

$=25[510]$

$=12750$

(v)
Integers from 1 and 500 which are multiples of 2 or 5 are

Multiples of 2 are 2, 4, 6, .....500. Therefore sum of all the multiples of 2 from 1 and 500 is

$a_{n}=500$

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

$\Rightarrow n=250$

$S_{250}=\frac{250}{2}(4+249 \times 2)=125 \times 502=62750$

and

Multiples of 5 are $5,10,15, \ldots ., 500$. Therefore sum of all the multiples of 5 from 1 and 500 is

$a_{n}=500$

$\Rightarrow 5+(n-1) 5=500$

$\Rightarrow n=100$

$S_{100}=\frac{100}{2}(10+99 \times 5)=50 \times 505=25250$

Multiples of both 2 and 5 are 10, 20, 30, ....., 500. Therefore sum of all the multiples of 2 and 5 from 1 and 500 is

$a_{n}=500$

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

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

$\Rightarrow n=50$

Now,

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

$S_{50}=\frac{50}{2}[20+(49) 10]$

$=25[510]$

$=12750$

Hence, the sum of Integers from 1 and 500 which are multiples of 2 or 5 is
[ sum of all the multiples of 2 + sum of all the multiples of 5 ] − [ sum of all the multiples of 2 and 5 ]
= [ 62750 + 25250 ] − 12750
= 75250