# Arithmetic sequencesWatch

#1
3 + 7 + 11 +.... Sum = 820

Find how many terms of the given arithmetic series must be taken to reach the given sum.
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9 years ago
#2
Use the formula for the sum of an arithmetic sequence.

This should be in your formula book

Sn= n/2(2a+(n-1)d)
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9 years ago
#3
What's the formula for the sum of the first n terms?

What's the first term and the common difference?
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9 years ago
#4
Use:

Sn = n/2[2a + (n-1)d]

Where d = common difference
a = first term
and Sn = Sum

Then, aim to find n.

Edit: Still not up to scratch on how to use LaTex. Sorry folks.
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9 years ago
#5
(Original post by Champagne Supernova)
Use:

Sn = n/2[2a + (n-1)d]

Where d = common difference
a = first term
and Sn = Sum

Then, aim to find n.
That.
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#6
I get 820 = n + 2n^2
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9 years ago
#7
Sounds like a quadratic to me... You know what to do?
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9 years ago
#8
(Original post by Flower_girl_16)
I get 820 = n + 2n^2
Work from here:

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9 years ago
#9
(Original post by bob9001)
Work from here:

Why 5n?
The difference is 4, so d = 4, therefore, it'd b 4n -4, no?
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9 years ago
#10

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#11
So 820 = n/2 (6 + 4n - 4)
820 = n/2 (4n + 2)
1640 = n (4n + 2)
1640 =4n^2 + 2n??

Is that not the same as what i said before??
820 = n + 2n^2

CS do i factorise?
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9 years ago
#12
(Original post by Flower_girl_16)
So 820 = n/2 (6 + 4n - 4)
820 = n/2 (4n + 2)
1640 = n (4n + 2)
1640 =4n^2 + 2n??

Is that not the same as what i said before??
820 = n + 2n^2

CS do i factorise?
Yep, that's right.
And, yes factorise, using his above method.
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#13
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9 years ago
#14
No worries. Oh, and welcome to TSR
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9 years ago
#15
Yeah factorise it.

Bob9001 it's 820 = n/2 (6 + 4n - 4)

Argh bloody slow TSR these days
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9 years ago
#16
Woops

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