Question

Find the sum of all integers between 1900 and 3000, inclusive, which are not divisible by 7.

Answer 1

I know its going to be multiples of 7 starting from 1904 and ending with 2996

Answer 2

See, first find all the sum of numbers from 1900 and 3000..

Answer 3

1900,1901,1902,1903,..........................3000
\(a = 1900\)
\(d = 1\)
\(l = 3000\)
Find number of terms ie \(n\)..

Answer 4

\[a + (n-1)d = 3000 \\ 1900 + n-1 = 3000 \\ n = 3001 - 1900\]

Answer 5

What is \(n\) equal to?

Answer 6

1101

Answer 7

Good..

Answer 8

To find sum, use:
\[Sum = \frac{n}{2}[a + l]\]

Answer 9

Find sum now..

Answer 10

2697450

Answer 11

Sure?

Answer 12

Okay, good..!!

Answer 13

Coming to second part..

Answer 14

Find sequence now for numbers divisible by 7 in that range..

Answer 15

1904, 1911, 1918.......................2996..

Answer 16

\(a = 1904, d = 7, l = 2996\)

Answer 17

Find \(n\) here..

Answer 18

Then find the sum and at last, subtract this sum from the sum you found earlier, all easy to do now?

Answer 19

sorry don't mean to interrupt
but I think your first answer to @eta 's question has the thousands digit number off

Answer 20

Which answer @freckles ?

Answer 21

his first answer

Answer 22

What it is?

Answer 23

the 2697450 answer
I have 2694450

Answer 24

Nope.. He is all good..

Answer 25

no i'm wrong sorry

Answer 26

\[Sum = \frac{1101}{2} \times 4900\]

Answer 27

i made a type-o in one of my numbers

Answer 28

It is okay, nothing to worry, at times I am also like you.. :P

Answer 29

Too bad none of us can be gods :p

Answer 30

There is not need too.. :P

Answer 31

I think @imyint you can do it now on your own..

Answer 32

yes thank you so much @eta !!!

Answer 33

If you come across any doubt, then ask @freckles as I am now going to sleep.. :P

Answer 34

Okay, good..
\(\dagger\)