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maths a level; sigma arithmetic problem

30
(4r+7)
r=3

Hello, yes i know it's quite late.

I'm very confused on how to solve this? i tried solving it the normal way i would if r=1, except by multiplying by 3 first. but the markscheme said it was wrong. (the answer should be 2044) how do i solve this? please explain.
Write out the first few terms to get the initial term and the difference term, then use the sum of an arithmetical series, which you can derive easily:

S=a+(a+d)+..(a+(n1)d)S = a + (a+d) + .. (a+(n-1)d)
S=(a+(n1)d)+(a+(n2)d)+..aS = (a+(n-1)d) + (a+(n-2)d) + .. a
2S=n[2a+(n1)d]\therefore 2S = n[2a+(n-1)d]
S=n2[2a+(n1)d]\therefore S = \frac{n}{2}[2a+(n-1)d]

What values of a, d and n did you get?
Original post by RogerOxon
Write out the first few terms to get the initial term and the difference term, then use the sum of an arithmetical series, which you can derive easily:

S=a+(a+d)+..(a+(n1)d)S = a + (a+d) + .. (a+(n-1)d)
S=(a+(n1)d)+(a+(n2)d)+..aS = (a+(n-1)d) + (a+(n-2)d) + .. a
2S=n[2a+(n1)d]\therefore 2S = n[2a+(n-1)d]
S=n2[2a+(n1)d]\therefore S = \frac{n}{2}[2a+(n-1)d]

What values of a, d and n did you get?


a = 19
d=4
n=30

is that correct??

i tried using that very same formula before, but to no avail.
Original post by ZiggyStarDust_
a = 19
d=4
n=30

is that correct??

Check your calculation of n.

However, I don't get the answer that you gave :frown:
Original post by RogerOxon
Check your calculation of n.

However, I don't get the answer that you gave :frown:


im worried that my markscheme might be wrong?? 2044 is apparently the answer.

So N is not 20? i don't understand what n could be then?
Original post by ZiggyStarDust_
30
(4r+7)
r=3

Hello, yes i know it's quite late.

I'm very confused on how to solve this? i tried solving it the normal way i would if r=1, except by multiplying by 3 first. but the markscheme said it was wrong. (the answer should be 2044) how do i solve this? please explain.


r=330=r=130r=12 \sum_{r=3}^{30}=\sum_{r=1}^{30} -\sum_{r=1}^{2}

because you'd have (u1+u2+u3+u4++u3)(u1+u2) (u_1+u_2+u_3+u_4+\cdots +u_3) - (u_1 + u_2)

can you see you'll be left with u_3 up to u_30 ?
r=3..30. If r were 1..30, you'd have 30 terms, so how many do you have for 3..30?
Original post by ZiggyStarDust_
im worried that my markscheme might be wrong?? 2044 is apparently the answer.

Sorry - I made a mistake - it is 2044.
Original post by rogeroxon
r=3..30. If r were 1..30, you'd have 30 terms, so how many do you have for 3..30?


o shet its 27 isnt it
Original post by ZiggyStarDust_
im worried that my markscheme might be wrong?? 2044 is apparently the answer.

So N is not 20? i don't understand what n could be then?


n=28 since the first term is 3 and the last term is 30 (so 28 terms all together).

Now solve, the answer should be 2044.
Original post by tripplea
n=28 since the first term is 3 and the last term is 30 (so 28 terms all together).

Now solve, the answer should be 2044.


why would it be 28 as a pose to 27?

because 30-3 is 27 right? so
Original post by ZiggyStarDust_
why would it be 28 as a pose to 27?

because 30-3 is 27 right? so


If
r=1-30 has 30 terms
then
r = 2-30 has 29 terms
r = 3-30 has 28 terms
Original post by tripplea
If
r=1-30 has 30 terms
then
r = 2-30 has 29 terms
r = 3-30 has 28 terms


thank you thank you thank you

so much

you've been very kind and helpful

its very much appreciated
you saved my arse

thank you
30
(4r+7)
r=3

19, 23, 27 - therefore the difference is 4
a=19, d=4, n=28

Sn = n/2[2a+(n-1)d]
28/2[2(19)+(28-1)4)
14[38+(27)4]
Sn = 2044

Hope that's clear enough for you :biggrin:
Original post by ZiggyStarDust_
thank you thank you thank you

so much

you've been very kind and helpful

its very much appreciated
you saved my arse

thank you


You're welcome. Do you get it now?

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