Question

find the sum of all numbers from 100 to 350 which are divisible by 4 and 6​

Answer 1

Given:

• A range of numbers 100 to 350.
• Two numbers 4 and 6.

To Find:

• The sum of all the numbers which are divisible by 4 and 6.

Answer:

Here we are given two numbers 4 & 6 and we have to find sum of numbers which are divisible by both of them between 100 and 350.

Now , Here's a concept :

The numbers which are divisible by both 4 and 6 will also be divisible by their LCM.

So ,LCM of 4 and 6 is 12 .

Hence ,we can find the sum of numbers which are divisible by 12 between 100 - 350 which will be equal to the sum of all numbers from 100 to 350 which are divisible by 4 and 6.

And ,

• First number after 100 which is divisible by 12 is 108.
• Last number which is divisible before 350 is 348.

Let us find which term is 348 .

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

$\sf{\implies 348= 108 +(n-1)12}$

$\sf{\implies 348-108=(n-1)12}$

$\sf{\implies 240=(n-1)12}$

$\sf{\implies (n-1) = \dfrac{240}{12}}$

$\sf{\implies (n-1)=20}$

$\sf{\implies n=20+1}$

$\bf{\leadsto n=21}$

Now we can use the formula of AP to find the sum.

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

$\sf{\implies S_{n}=\dfrac{21}{2}[2\times108+(21-1)12]}$

$\sf{\implies S_{n}=\dfrac{21}{2}[216+20\times12]}$

$\sf{\implies S_{n}=\dfrac{21}{2}[216+240]}$

$\sf{\implies S_{n}=21(108+120)}$

$\sf{\implies S_{n}=21\times 228}$

$\bf{\implies S_{n}=4788}$

Hence the required sum is 4788.

Answer 2

Answer:

same as above !

actually i came here to see for the answer but got the opportunity to answer so i thought why not to take free points bcoz the above answer would be popular only and it is verified too so i will not try to write same in precise too

thank you stranger to give me this answer

actually looking for the solution to submit the homework!