Arithmetic series sum

[radnorgardens]

Hello all, would really appreciate your assistance with the following:

Question:
The sum of an arithmetic series is 270. The common difference is 1 and the first term is 4. Calculate the number of terms in the series.

Formula: S₂₀ = n/2(2a+(n-1)d
Where: n = number of terms, a is the first term, and d is the common difference.

So I know I need to isolate n by transposing the formule, but can't get the correct answer (I know it is 20). My attempt:

270 = n/2(2(4)+(n-1)1
270 = n/2(8+(n-1)1
270 x 2 = n(8+(n-1)1
540 = n(8+(n-1)1
540 x 1 = n(8+(n-1)
540 = n(8+(n-1)
540 = 8n+n^₂-n
540/8 = n+n^₂-n
67.5 = n2
8.23 = n

Thank you all.

 

[skeeter]

Hello all, would really appreciate your assistance with the following:

Question:
The sum of an arithmetic series is 270. The common difference is 1 and the first term is 4. Calculate the number of terms in the series.

Formula: S₂₀ = n/2[2a+(n-1)d]
Where: n = number of terms, a is the first term, and d is the common difference.

So I know I need to isolate n by transposing the formule, but can't get the correct answer (I know it is 20). My attempt:

270 = n/2[2(4)+(n-1)1]
270 = n/2[8+(n-1)1]
270 x 2 = n[8+(n-1)1]
540 = n[8+(n-1)1]
delete this unnecessary step ...
540 = n[8+(n-1)]
540 = n[n+7] ... correction

now continue ...

can you finish?

 

[Dale10101]

Method

Hello all, would really appreciate your assistance with the following:

Question:
The sum of an arithmetic series is 270. The common difference is 1 and the first term is 4. Calculate the number of terms in the series.

Formula: S₂₀ = n/2(2a+(n-1)d
Where: n = number of terms, a is the first term, and d is the common difference.

So I know I need to isolate n by transposing the formule, but can't get the correct answer (I know it is 20). My attempt:

270 = n/2(2(4)+(n-1)1
270 = n/2(8+(n-1)1
270 x 2 = n(8+(n-1)1
540 = n(8+(n-1)1
540 x 1 = n(8+(n-1)
540 = n(8+(n-1)
540 = 8n+n^₂-n
540/8 = n+n^₂-n
67.5 = n2
8.23 = n

Thank you all.

To help yourself, take 20 and plug it into the original formula to verify that it and the formula agree. You know, perhaps you wrote one or the other incorrectly. If they agree then plug 20 into one of your intermediate steps and work your way forward or backward to find which step goes from 270 = 270 to 270 = something else.

Note that even if you use your incorrect answer you can plug it into intermediate steps to at least verify that your steps going backwards are consistent with themselves.

What I mean is that since you are saying n = 8.23, then you also claim that n^2 = 67.5 and so on back up the chain until the right hand side of the equation doesn't match the left hand side of the equation if there is an error.

I didn't realize this simple truth for years and would spend 5 times as much time looking for my dumb errors than doing math.

 

[radnorgardens]

Thanks guys. Ok so from Skeeter's clarification 540 = n[n+7]. I can see that 540 = 20[20+7] becomes 20x27 = 540. So that's all good (thanks Dale). But it goes wrong from here. I've tried a few routes:

First:
540 = n[n+7]
540 = n²+7n
540/7 = n²+n
77.14 = n²+n
8.78 = 2n
4.39 = n

Second:
540 = n[n+7]
540-7 = n2
533 = n²
23.09 = n

Third:
540 = n x [n+7]
540/[n+7] = n

The third one feel like it's the right path (wrong?), but have no idea how to move it forward.

Again your help would be much appreciated.

 

[ksdhart]

I think you're making the process more complicated than it needs to be. You have the equation:

n[n+7] = 540
n^2+7n = 540
n^2+7n-540 = 0

Now that looks like an equation where you can use the quadratic formula (you can factor it if you like, but I personally prefer the quadratic formula). Try that and see what you get.

 

[relief]

a₁ + a₂ + a₃+ ... +a_n = 270

a₁ =4, d=1

n=?

---

We have so :

4+5+6 + ... + k = 270 (1)

Use a simple trick :

(1+2+3)+4+5+6+ ... + k - (1+2+3) = 270

1+2+3+4+5+6+ ... + k - 6 = 270

1+2+3+4+5+6+ ... + k =270+6

k(k+1)/2=276

k(k+1)=276 x 2

k(k+1) =552

k(k+1) = 23 x 24

k=23 (2)

Now, from (1) and (2), we have :

4+5+6+ ... +23 =270

a₁ =4, a_n=23 (consecutive integers)

n = 23 - 3 = 20

 

[radnorgardens]

Thank you all. I have covered quadratic equations, just didn't think to use it for answering this question!!! Use that, I did come up with n=-27 and n=20. Thank you for helping. I have another problem now, but I think I'll post that separately.