# Sequences

Find the sum of all the integers between 100 and 400 that are divisible by 7
To find n, should I minus 400 from 100 and then divide by 7?
I want to make sure if it’s the correct method
the series is 203+ 210+ ….+399

nth term = a +(n-1)d therefore 399 = 203+(n-1)7

196 =(n-1)7 which gives n=29

therefore sum =29 (203+399)/2 =8729
Original post by rosie992.625
the series is ...

Its best to just give hints, and note the series "starts" at 100
Original post by rosie992.625
the series is 203+ 210+ ….+399

nth term = a +(n-1)d therefore 399 = 203+(n-1)7

196 =(n-1)7 which gives n=29

therefore sum =29 (203+399)/2 =8729

n is 43
Original post by 90degrees
n is 43

There are 43 terms in the series, I agree. So what are you unsure about?
Original post by mqb2766
There are 43 terms in the series, I agree. So what are you unsure about?

What’s your method to find 43?
Original post by 90degrees
What’s your method to find 43?

Its probably worth having a read through the forum guidelines (sticky at the top of the forum), especially about posting your working ...

If you know the answer (it is a relatively simple calculation), post what you thought/tried and what you are unsure about. Preferably do this in the original post, so people can offer approriate help/hints.

* You don't say whether your sequence/series starts at index 0 or 1 which affects the value of n.
* You could try a simple example of the number of terms between 100 and 104, 100 and 111, 100 and 112, ... does your method give the right solution (what is the right solution for these simple cases)
* ...
(edited 1 year ago)
Original post by mqb2766
Its probably worth having a read through the forum guidelines (sticky at the top of the forum), especially about posting your working ...

If you know the answer (it is a relatively simple calculation), post what you thought/tried and what you are unsure about. Preferably do this in the original post, so people can offer approriate help/hints.

* You don't say whether your sequence/series starts at index 0 or 1 which affects the value of n.
* You could try a simple example of the number of terms between 100 and 104, 100 and 111, 100 and 112, ... does your method give the right solution (what is the right solution for these simple cases)
* ...

I found that a= 105 and l= 399 and im unsure about what’s the exact method to find n
Original post by 90degrees
I found that a= 105 and l= 399 and im unsure about what’s the exact method to find n

I agree with those. Its really a fence post / fence panel type problem which if you forget how to do it (you should have covered it several years ago), just use some simple examples.
* If a=105 and l=105, what is n?
* If a=105 and l=112, what is n?
...
you should be able to spot/justify the method.

Note here you;ve still not really defined what n is, but Im presuming its the number of terms in the sequence/series.
(edited 1 year ago)
Original post by mqb2766
I agree with those. Its really a fence post / fence panel type problem which if you forget how to do it (you should have covered it several years ago), just use some simple examples.
* If a=105 and l=105, what is n?
* If a=105 and l=112, what is n?
...
you should be able to spot/justify the method.

Note here you;ve still not really defined what n is, but Im presuming its the number of terms in the sequence/series.

Yes, n is the number of terms in a sequence
Original post by 90degrees
Yes, n is the number of terms in a sequence

So do you get the right answer?
I got the answer, I just need the correct method to find n
Original post by 90degrees
I got the answer, I just need the correct method to find n

It should be fairly self evident, but if you have a, l and d, just sub them into the arithmetic sequence formula and solve for n. Its what the "method" would do but a bit overkill perhaps.
(edited 1 year ago)
Original post by mqb2766
It should be fairly self evident, but if you have a, l and d, just sub them into the arithmetic sequence formula and solve for n. Its what the "method" would do but a bit overkill perhaps.

I understood, thank you