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## Comment on

Multiples of 5## X => 5, 10, 15,..........200

+200,195,190,........... 5

-------------------------------

205,205,205...........205

there are ( (200-5) / 5 ) + 1 = 40 numbers of that are multiples of 5 in 5,10,.....200 sequence.

now 40(205)/2 = 4100.

## Excellent!

Excellent!

## Can you please explain

## yogasuhas is using an

yogasuhas is using an approach described from 0:57 to 2:40 in the video on sums of sequences (https://www.greenlighttestprep.com/module/gre-word-problems/video/930)

yogasuhas determined that there are 40 terms from 5 to 200 inclusive.

Then yogasuhas added the same sequence, but in reverse order (200, 195, 190, ..., 10, 5)

Then we add BOTH sequences together, BUT we do this in PAIRS.

So, 5 + 200 = 205

10 + 195 = 205

15 + 190 = 205

.

.

.

185 + 20 = 205

190 + 15 = 205

195 + 10 = 205

200 + 5 = 205

The result of adding 40 PAIRS of values is a sum that equals 40 x 205

However, since we added every term TWICE, we must divide our sum by 2.

We get: (40 x 205)/2

Does that help?

Cheers,

Brent

## This was actually a hard

I tried the bunching method of 5 + 200, 10 + 195, etc. to see that you would have to multiple 205 by the number of pairs of numbers, but I had no idea had to find how many pairs there are. What's the equation for that? You would literally have to write out all the pairs. I could have found the answer, but there has to be some kind of shortcut method to finding pairs.

Anyway, I like the factoring thing you did. I think that's the best method to use when you're asked to find the sum of MULTIPLES of a number. Gonna write that down on my sneaky cheat sheet for the at-home exam!! Thank you!

## "I had no idea had to find

"I had no idea had to find how many pairs there are. What's the equation for that?"

We want to add MULTIPLES of 5 from 5 to 200 inclusive.

5 = 5(1)

10 = 5(2)

15 = 5(3)

.

.

.

195 = 5(39)

200 = 5(40)

So the number of multiples of 5 from 5 to 200 inclusive = the number of integers from 1 to 40 inclusive

From here we might just recognize that there are 40 such integers.

Or we can apply the following formula:

The number of integers from x to y inclusive equals y - x + 1

Cheers,

Brent

## Is it not possible to

So, 102.5 x 41 = 4,202.5.

Where did I go wrong / misstep on this method?

## Is that approach will work.

Is that approach will work. The only thing you missed it is. There are 40 terms (not 41).

So the sum = (40)(102.5) = 4100