Need maths help

Hi guys, I get how to do all the parts for this question except for:
iii) I know the extended table way but the calculation via logarithms way? iv) Again, I know the extended table way but the direct calculation way bit in the question is confusing?
I've attached the full question and markscheme.
Will appreciate a lot any help.

Taking (iii) for example, scheme B has a geo sequence of $a_n = 20 \cdot 1.12^{n-1}$ for payment in the nth month. For total, we take the sum of this from the first month up to an arbitrary month $k$, which is $\displaystyle \sum_{n=1}^k a_n = \dfrac{500}{3}(1.12^k-1)$

We are interested when this sum reaches 1000. Hence we do $1000=\dfrac{500}{3}(1.12^k-1)$ and use logs to determine the month $k$. In problems like these, you'd like to take $\lceil k \rceil$ (which means the smallest integer greater than or equal to $k$) as the answer because the price here increases every month, and not continuously.

Thanks so much for your reply! How did you get 500/3(1.12^k-1)?
Many thanks

Summing up a geometric series. You should know what the formula is if you covered it.

Thank you so much!! I get iii).
For iv) I did
let's say the total paid by both scheme A and B is equal in month x.
Scheme b:
Sx = 500/3(1.12^x -1)
Scheme a:
Sx = (1/2)x(4x+36)

and then I did..
500/3(1.12^x -1) = (1/2)x(4x+36)

And tried to solve it for x but then it got really messy..
ended up with 2x^2 +18x -500/3(1.12^x) +500/3=0

Can you please help me with iv)? Many thanks for all your help

You can't solve that for $x$, it's a transcendental equation.
The best you can do with this approach is set up an iterative scheme and approximate $x$

Thanks so much, but what do you mean by transcendental?
Also, with the iterative scheme do you mean what is shown in the mark scheme for this question?

https://en.wikipedia.org/wiki/Transcendental_equation

No, by iterative scheme I mean that you need to rearrange for one of the k's so that $k=f(k)$ and define a sequence $k_{n+1}=f(k_n)$ with an appropriate starting point $k_0$ then iterate, but this is probably above and beyond C1/C2, especially without any guidance on the set-up. Hence the table is easier to work with on this level of knowledge, so just work off that instead for this question.

Is this GCSE or A level?

Taking (iii) for example, scheme B has a geo sequence of $a_n = 20 \cdot 1.12^{n-1}$ for payment in the nth month. For total, we take the sum of this from the first month up to an arbitrary month $k$, which is $\displaystyle \sum_{n=1}^k a_n = \dfrac{500}{3}(1.12^k-1)$

We are interested when this sum reaches 1000. Hence we do $1000=\dfrac{500}{3}(1.12^k-1)$ and use logs to determine the month $k$. In problems like these, you'd like to take $\lceil k \rceil$ (which means the smallest integer greater than or equal to $k$) as the answer because the price here increases every month, and not continuously.

Hi RDKGames, just randomly attempted this question -iii)why would you take the smallest integer? I got a value of k is greater than or equal to 18.309 so shouldn't scheme A be 19 like it says in the mark scheme? I got a value of k is greater than or equal to 17.17 so shouldn't scheme B be 18 like it says in the mark scheme?

Yes that's correct. You are doing exactly what I said in my post. What's confusing you??

In problems like these, you'd like to take $\lceil k \rceil$ (which means the smallest integer greater than or equal to $k$) as the answer because the price here increases every month, and not continuously.

Oh it's because I thought you meant for e.g. scheme B : k≥ 17.17 , you'd have to take 17 as the answer? Highlighted where in your post implies that.

But np - I thought maybe you were referring to the mark scheme being wrong.

The highlighted part does not imply that. It means that you take the next integer above your number, which is precisely 18 for 17.17. So $\lceil 17.17 \rceil = 18$.