Leaving Cert 09 Maths OL Paper 1 Q5 C(iii) - Quick Question!

Trotter

Hi all, I'm trying to give a friend some help with the leaving cert papers from 2009.

In Q5 Paper 1, Sequences and Series -

The first two terms of a geometric series are − 6 +12 + ….
(i) Find r, the common ratio.
(ii) Find 7 T , the seventh term of the series.
(iii) Starting with the first term, how many terms of the series must be added to give a sum of 30.


I'm having trouble explaining part (iii).

I make the common ratio r to be -2 from putting T2 over T1 (+12 over -6) and thats confirmed in the answers provided to examiners.

However, in that examiners document, for part (iii) it uses the formula

Sn= a(1-r^n) over 1-r

According to my maths revision book here, the above formula is only for
(-1 < r < 1) so -2 doesnt seem to fit in here.

I'd have used the version

Sn = a(r^n-1) over r-1 which is for (r>1 or r< -1)

because r = -2 fits the second formula.




What am I missing because I can't figure out how formula 1 was used when r was -2.

Clinker

The formula [latex]$\displaystyle S_n = \frac{a(1 - r^n)}{1 - r}$[/latex] is used for the sum of the first [latex]$n$[/latex] terms of a series. This is finite, so the formula works for any value of [latex]$r$[/latex]. What you're thinking of is the formula [latex]$\displaystyle S_{\infty} = \frac{a}{1 - r}$[/latex], which gives you the sum of the infinite series. This indeed only works when [latex]$- 1$[/latex] < r < 1.

Trotter

Hmm.. those formulas I have there are straight from the Less Success More Stress book for ordinary level maths. They dont use the infinite formula at ordinary level I think.

At least I know now that 1-r^n formula works for all values of n. Thanks for that!

Fringe

Don't think the -1 < r < 1 is relevant here. It should only apply for infinite series.

Also, note that a.(r^n - 1)/r - 1 = a(1 - r^n)/1 - r
Think of it as multiplying by -1/-1 which is simply just 1. Same thing as why -3/-4 = 3/4. So for this expression, any value for r except for 1 will work.

Trotter

Gotcha. Thanks for that. I'm a primary teacher so its a while since I had to remember this stuff.

MathsManiac

As Fringe points out, the two versions of the formula that you refer to are actually the same.

The only reason that some people prefer to have two versions is so that, when they stick the numbers in, they'd rather have a positive number above and below than have a negative number above and below.

Since either version will work for all values of r (other than r=1), then, if teaching an ordinary level LC student, I would be inclined to stick with the one version the whole time - the one with the "1-r" underneath. It's the one that's in the "Formulae and Tables" book that they will have available in the examination, (page 22).

dustintheturkey

Hi,
the reason for the modulus of r must be less than 1, is that when you are summing an infinite series of numbers, the sum must "converge" to a certain value.
Imagine the sum being "y" on an x-y axis, the curve must "bend" downward as it's going up.
Otherwise the sum will "diverge", bending in the wrong direction to plus or minus infinity as you keep adding the next term.

Yes, the two formulae are also the same. One is (-1)/(-1) times the other for "convenience".

It's much more fun working the formula out though! good practice.