Question

find the sum of all Integers between 100and 200 that are divisible by 4​

Answer 1

Answer:

3900

Step-by-step explanation:

The numbers 104,108,112,.......200 are divisible by 4

a = first term = 104

d = common difference = 4

nth term = 196

196=104+(n-1)4

196=104+4n-4

196=100+4n

96=4n

n=24

Sum of n terms of an ap = n/2(a+l)

Sum of n terms of an ap = 12(104+196)

Sum of n terms of an ap = 12(300)

Sum of n terms of an ap = 3600

3600 is the sum of all integers divisible by 4

Formulae used :

nth term of an A.P = a+(n-1)d

Sum of n terms of an A.P = n/2(2a+(n-1)d) or n/2(a+l)

Answer 2

Given:

• A interval of 100-200.
• A number 4 is given to us.

To Find:

• Sum of all integers between 100 & 200 which are divisible by 4 .

Concept Used:

• We will make use of Arithmetic Progression to find the sum.

Answer:

We are required to find the sum of numbers between 100 & 104 which are divisible by 4.

Now the first integer after 100 which is divisible by 4 is 104.

And the last number before 200 divisible by 4 is 196.

Now

The required series will be 104,108,112...........196.

And ,

Sum of series will be , 104+108+112+....196.

Here ,

• First term is 104 .
• Last term is 196.

So we must find which term is 196 ,in order to find sum of the series.

Formula of nth term is :

$\large{\boxed{\red{\sf{\leadsto T_{n}=a+(n-1)d}}}}$

using this ,

$\sf{T_{n}=a+(n-1)d}$

$\sf{196=104+(n-1)4}$

$\sf{196-104=(n-1)4}$

$\sf{(n-1)4=92}$

$\sf{(n-1)=\dfrac{92}{4}}$

$\sf{(n-1)=23}$

$\sf{n=23+1}$

${\underline{\boxed{\red{\sf{\longmapsto n=24}}}}}$

Hence total number of terms is 24 .

$\rule{200}1$

Now Formula for sum of n terms of an AP is given by ,

$\large{\boxed{\red{\sf{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}}}}$

Here

• First term is 104.
• Common difference is 4 .
• Number of terms is 24.

$\sf{S_{n}=\dfrac{24}{2}[2\times 104+(24-1)4]}$

$\sf{S_{n}=12[208+23\times4]}$

$\sf{S_{n}=12[208+92]}$

$\sf{S_{n}=12\times 300}$

${\underline {\boxed{\red{\sf{\leadsto S_{n}=3600}}}}}$

Hence sum of series is 3600.