arithmetic progression

[nitisha beezloll]

The sum to n terms of a particular series is given by S = 17n - 3n^2.
Find an expression for the nth term of the series.
pls help me out!

 

[tkhunny]

The terminology is suppose to mean something to you. Seek the defintiions of things.

It's an arithmetic progression.

First term: a
Second term: a + d
third term a + 2d
etc.

Add up 'n' of those and see what you get.

 

[mmm4444bot]

pls help me out!

Where are you stuck, in this exercise? Please be specific.

You could try the following strategy, if you already understand arithmetic sequences and nth-partial sums.

(If you do not yet understand these two concepts, then you're not ready for this exercise; go back and review the basics for arithmetic sequences and their partial sums.)

Use the given formula to calculate the first two partial sums.

Clearly, the first partial sum is simply the first term: a[sub:v6013q02]1[/sub:v6013q02]

Subtracting the first partial sum from the second yields the second term in the sequence: a[sub:v6013q02]2[/sub:v6013q02]

Once you have these, you can easily determine the sequence's common difference.

Now you're ready to plug the required values into the famous formula for the nth term:

a[sub:v6013q02]n[/sub:v6013q02] = a[sub:v6013q02]1[/sub:v6013q02] + (n - 1)(d)


Simplify your result, if that's what you've been doing in class; otherwise, you can just leave it in the form above.

Cheers ~ Mark

 

[soroban]

Hello, nitisha beezloll!

I already solved this problem at another site.
For those of you who just tuned in, here's my solution . . .

\(\displaystyle \text{The sum to }n\text{ terms of a particular arithmetic series is given by: }\: S \:=\: 17n - 3n^2\)

\(\displaystyle \text{Find an expression for the }n^{th}\text{ term of the series.}\)

\(\displaystyle \text{We have: }\:S \:=\:17n - 3n^2,\,\text{ a polynomial in }n.\) .[1]


\(\displaystyle \text{The sum of an arithmetic series is: }\:S \:=\:\tfrac{n}{2}\left[2a + (n-1)d\right]\)

. . \(\displaystyle \text{which simplifies to: }\:S \:=\:\left(a - \tfrac{1}{2}d\right)n + \left(\tfrac{1}{2}d\right)n^2,\,\text{ a polynomial in }n.\) .[2]


In [2] and [1], equate coefficients: .\(\displaystyle \begin{Bmatrix}a-\frac{1}{2}d &=& 17 & [3] \\ \frac{1}{2}d &=& \text{-}3 & [4] \end{Bmatrix}\)

\(\displaystyle \text{From [4]: }\:\tfrac{1}{2}d \:=\:\text{-}3 \quad\Rightarrow\quad d \:=\:\text{-}6\;\)

\(\displaystyle \text{Substitute into [3]: }\:a - \tfrac{1}{2}(\text{-}6) \:=\:17 \quad\Rightarrow\quad a \:=\:14\)

\(\displaystyle \text{Therefore: }\:a_n \;=\;14 + (n-1)(\text{-}6) \quad\Rightarrow\quad \boxed{a_n \:=\:20 - 6n}\)

 

[Denis]

Soroban, all jokes about "doing homework" aside,
how do you figure this'll help the student?

 

[soroban]

It will show him/her/everyone an alternate approach to the problem.

. . (I bet this method didn't occur to most of y'all.)

 

[nitisha beezloll]

Thank you to all for the help...thnk ya mr sorobun!! it really helped...