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Arithmetic Sequences 1

Just been doing some revision - and noticed that arithmetic sequences is a weak point.

I've been doing a couple of the solomon C1 papers and this was one of the questions:

Evaluate

30
∑ (7 + 2r).
R=10

So I did the following:

Worked out a couple of numbers at the beginning of the sequence:

27 + 29 + 31 +33....67 a=27 d=2

Popped those numbers into the equation:

Sn = n/2 (2a + (n-2)d)

sn = 30/2 (2 x 27 +(30-1)2)

I got 1,680 - this is the wrong answer according to the mark scheme...

This is the link to the answer - and this question is question 2 in paper A - I am not quite sure where the 21 comes from....??

https://a9497d7f220e174d38d014b2b1d86a68bedc42a8.googledrive.com/host/0B1ZiqBksUHNYNExoNEtSS1ViN2c/for-Edexcel/Solomon%20A%20MS%20-%20C1%20Edexcel.pdf

Reply 1

Zacken

Original post by christinajane

...

The formula you used is only for the sum of the first n terms. i.e: it needs to start from the first term.

Your sum, on the other hand is:

\displaystyle \sum_{r=10}^{30} (7+2r)

which starts from the

r=10

th term.

To fix this, you can write your sum as:

\displaystyle \sum_{r=1}^{30} (7+2r) - \sum_{r=1}^{9} (7+2r)

i.e: you sum the first

30

terms allowing you to use the formula and then subtract the sum of the first

9

terms (with the formula) to compensate.

Reply 2

christinajane

Original post by Zacken

The formula you used is only for the sum of the first n terms. i.e: it needs to start from the first term.

Your sum, on the other hand is:

\displaystyle \sum_{r=10}^{30} (7+2r)

which starts from the

r=10

th term.

To fix this, you can write your sum as:

\displaystyle \sum_{r=1}^{30} (7+2r) - \sum_{r=1}^{9} (7+2r)

i.e: you sum the first

30

terms allowing you to use the formula and then subtract the sum of the first

9

terms (with the formula) to compensate.

Ahhh yeah! Thats a tricky one - No wonder I am so bad at them - I thought you could just plug the numbers in regardless of what term n starts at. Now I know where 21 comes from!

I never realised it had to start at 1 - as stupid as that sounds.Thanks again zacken!

You're the best!

Reply 3

Zacken

Original post by christinajane

No worries, good on you for clearing it up now and making an effort to understand! :smile:

Reply 4

Kevin De Bruyne

Original post by christinajane

Just to add, you could do what you did right to the point where you choose n.

Instead of 30 terms, the 1st term is the 10th term, the 2nd term is the 11th term... so the 30th and final term (represented by the 30 on top) is the 21st term, so you can say it is the sum of 21 terms with a = 27 and d=2, like you did. It's just more rigorous to take the sum from 1 to 9 away from the sum of 1 to 30 to give the sum of 10 to 30.

Reply 5

Zacken

Original post by SeanFM

Instead of 30 terms, the 1st term is the 10th term, the 2nd term is the 11th term... so the 30th and final term (represented by the 30 on top) is the 21st term, so you can say it is the sum of 21 terms with a = 27 and d=2, like you did.

Just to clarify, is this defining a new 'shifted' arithmetic sequence? If so, that's rather neat and just as rigorous as my method. I've never really thought about doing this explicitly before!

Reply 6

Kevin De Bruyne

Original post by Zacken

Just to clarify, is this defining a new 'shifted' arithmetic sequence? If so, that's rather neat and just as rigorous as my method. I've never really thought about doing this explicitly before!

Yes, nor have I. :rofl:

I think it would work out in the exam as long as you explain what you're doing.

Reply 7

christinajane

Original post by Zacken

Just to clarify, is this defining a new 'shifted' arithmetic sequence? If so, that's rather neat and just as rigorous as my method. I've never really thought about doing this explicitly before!

It all makes sense when its explained to me properly - like now I understand why my answer was too high for this question -

if like me I am doing the sum from one right up to 30 - which is essentially what I did - instead of 21 terms.

Another thing that catches me out is - is when to count the first n term

uhm like in this case it is 30 - 9 = 21

I probably still would have done 30 - 10 if I hadnt of seen the answer.

Gets me every time - that n-1 one thing and when to apply it.....

Reply 8

Kevin De Bruyne

Original post by christinajane

It's just something that you have to get used to. If you have the numbers 10, 11, 12.. all the way up to 30, there are 21 numbers there and so 21 terms.

You could count with your hand - 10 is the first term, 11 is the second, 12... and you find that your 10th digit will be the 19th term, so the 20th will be the 29th term, so the 30th term will be 21. :h: