# Solve Series P: Sum of First n Terms & Arithmetic Progression

• ings
From the forum's FAQ:How should I ask for help with my homework?Do not post full problem solutions (i.e. you may post solutions to sub-parts of the problem). Do not ask for confirmation of your solution, just ask for help with the problem.In summary, the conversation discusses finding the sum of the first n terms of a series P, which has a sum of 20n-4n2 for the first 2n terms. The conversation explores the assumption that the series is an arithmetic series and deduces the expression for Sn, the sum of the first n terms. The conversation also considers other possible series and concludes that the original assumption of an arithmetic series is correct.
ings
CAN U ALL HELP ME TO SOLVE THIS QUESTION?
I don't know how to start...

The sum of the first 2n terms of a series P is 20n-4n2. Find in terms of n, the sum of the first n terms of this series. Show that the series is an arithmetic series.

Well if the sum of 2n terms is ... then what they gave you is equal to that. Now the formula for the sum of n terms is ... and you have to substitute that into the sum of 2n terms formula by manipulating a few things.

But the question only mentions that sentences ...

Hmm... now that I read it again I was planning on assuming the series is arithmetic. But even then I can't seem to isolate Sn without having d in the equation. Which class are you in?

If I assume it is an arithmetic series the series is uniquely determined.

But I don't think the statement above is enough. I could easily fix an arbitrary series of number that give the same sum for the first 2n terms. We have two numbers to choose ($$a_{2n+1}, a_{2n+2}$$) and only one constraint ($$a_{2n+1}+a_{2n+2}=16-8n=S_{2n+2}-S_{2n}$$).

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ings said:
CAN U ALL HELP ME TO SOLVE THIS QUESTION?
I don't know how to start...

The sum of the first 2n terms of a series P is 20n-4n2. Find in terms of n, the sum of the first n terms of this series. Show that the series is an arithmetic series.
So the "0" th term is 0, the sum of the first 2 terms is 20(1)- 4(1)= 16, the sum of the first 4 terms is 20(2)- 4(4)= 24, the sum of the first 6 terms is 20(3)- 4(9)= 24, the sum of the first 8 terms is 20(4)- 4(16)= 16, the sum of the first 10 terms is 20(5)- 4(25)= 0, etc. In order to go up from 0 to 16 in two steps, the series must have a positive difference. But after the 6th term, it is decreasing so must have a negative difference. I don't see how that can be an arithmetic sequence that must have a constant difference.

I'm pretty sure that this is a basic "school type" problem that *assumes* that all series are either Arithmetic or Geometric. A geometric series will have a sum that is an exponential function of "n" and an arithmetic series will have a sum that's a quadratic function of "n". So out of those two options it's clearly Arithmetic.

In terms of the first term "a" and the common difference "d" the sum of an AS is :

S(n) = (a-d/2) n + d/2 n^2

so S(2n) = (2a-d) n + 2d n^2, and the rest is very straight foward.

You don't need to assume anything about the type of sequence. The key is to find the expression for Sn first, which you can deduce from S2n, and then calculate an=Sn-Sn-1. You'll find that an is an arithmetic sequence.

I don't think you can deduce $$S_n$$
I can do arbitrary series to get the result, e.g.
$$a_n=(1+(-1)^n)(12-2n)$$
will give the requested sum, but is not an arithmetic series, as every odd term is 0.
The first terms in this series are
$$a_1=0,a_2=16,a_3=0,a_4=8,a_5=0,a_6=0,a_7=0,a_8=-8,a_9=0$$
And the sum of the first 2n terms is given by the formula above.

I guess it depends on whether you interpret "the sum of the first 2n terms" to mean S2n, in which case, it's straightforward to find Sn, or you interpret it to mean that f(n)=20n-4n2 happens to equal S2n for n=0,1,...

I don't get the difference on that

It's the difference between figuring out f(x) if you know what f(2x) as opposed to figuring out f(x) if all you know is f(x)=g(x) for x=0, 1, 2, ...

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i am not sure whether it's correct...

i do it in this way: let S2n= f(2n)
then Sn=f(n)

f(2n)= 20n-4n2
= 10(2n)- (2n)2

hence, f(n)= 10n - n2
SO Sn = 10n - n2

Is it correct?

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ings said:
i am not sure whether it's correct...

i do it in this way: let S2n= f(2n)
then Sn=f(n)

f(2n)= 20n-4n2
= 10(2n)- (2n)2

hence, f(n)= 10n - n2
SO Sn = 10n - n2

Is it correct?
Yes, that looks good. Next, try using vela's advice in Post #8 to find an.

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## 1. What is the formula for finding the sum of the first n terms of an arithmetic progression?

The formula for finding the sum of the first n terms of an arithmetic progression is Sn = (n/2)(2a + (n-1)d), where n is the number of terms, a is the first term, and d is the common difference.

## 2. How do you determine if a given series is an arithmetic progression?

To determine if a given series is an arithmetic progression, you can check if the difference between consecutive terms is constant. If the difference is the same for all pairs of terms, then the series is an arithmetic progression.

## 3. Can the sum of the first n terms of an arithmetic progression be negative?

Yes, the sum of the first n terms of an arithmetic progression can be negative if the common difference d is negative and the number of terms n is odd.

## 4. How does the sum of the first n terms of an arithmetic progression change if the common difference is doubled?

If the common difference is doubled, the sum of the first n terms of an arithmetic progression will also be doubled.

## 5. Can the sum of the first n terms of an arithmetic progression be infinite?

No, the sum of the first n terms of an arithmetic progression can never be infinite as long as the common difference is a finite number.

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