having difficulties with Geometric series question

The problem I have is with the worked solutions is how did the author come up with:
$\frac{1}{a}(\frac{r^{n-1}+r^{n-1}+r^{n-2}+r^{n-31}+...+r^{0}}{r^{(n-1)(n-2)(n-3)..3.2.1}})$

I thought that $\frac{1}{a}+\frac{1}{ar} + ....+ \frac{1}{ar^{n-1}} = \frac{1}{a}(r^{0}+r^{-1}+r^{-2}+ r^{-3})$
therefore r = common ratio = $r^{-1}$

and lastly how did the author need up with $\frac{a^{n-1}n(n-1)}{2P_{n}}$ where $P_{n} = a^{n}\frac{r^{n(n-1)/2}}{2}$ as the answer

I thought $P_{n} = a^{n}\frac{r^{n(n-1)/2}}{2}$ is wrong.
Shouldn't $P_{n} = a^{n}r^{\frac{n(n-1)}{2}}$ ?
Please if someone can be kind to explain to me thank you

No idea - just looks wrong.



I assume you meant that to continue, rather than stop at $r^{-3}$

If so, then agreed.

Though in your subsequent working you should have 1-(1/r) in the denominator, not 1-r.



Looks like another error in the book

its the same book i showed you in the other thread

I thought it was the same volume.

I stand by my previous post.

Sorry i want to ask are you familiar with AEB questions?

Difficult to know, I don't associate any particular questions with AEB - even had to look up what the abbreviation meant.

AEB - associated examining board

I wanted to ask about questions like this:
The ninth term of an arithmetic progression is 52 and the sum of the first twelve terms is 414. Find the first term and common difference. The rth term of an arithmetic progression (1+4r). Find, in terms of n, the sum of the first n terms of the progression.

The point i wanted to make is the The rth term of an arithmetic progression (1+4r). part. From your opinion do you think that this part is added to distract or confused students.

I thought it was the same volume.

I stand by my previous post.

I would take this to be two different questions, or two parts of a question dealing with two different series.

From the way you've written it, it looks to be one, but that may just be how you've written it.

Can you link/scan/photo the original?

number Q 25 thank you

To be brutally honest, it doesn't look like the author should be writing books about mathematics!

Agreed. Out of interest I tracked down the publishers. It's a private limited company with two shareholders. No prizes for guessing who the company director is!

Didn't realise vanity publishing extended to the academic sector - perhaps I'm being a bit mean there....

To be brutally honest, it doesn't look like the author should be writing books about mathematics!