# How much gold can I buy with $2000$ cash, if the price starts at $20$ and increments by $0.01$ with each purchase?

I have problem with arithmetic.

For example,

\begin{align} \text{My cash} &= 2000 \\ \text{Gold Price} &= 20 / \text{gold} \\ \text{Increment / buying gold} &= 0.01 \end{align}

I would like to buy gold using my cash. How to get the actual gold amount that I can buy?

Just for enlightment

For example, I would like to buy 5 gold, so it would be as follows

$$20+20.01+20.02+20.03+20.04$$

Total price to spend would be,

\begin{align} u_n &= \text{initial}+(n-1)\cdot\text{diff} \\ &= 20+(5-1)\cdot 0.01 \\ &= 20.04 \\ \\ \text{sum} &= \frac{n}{2}(\text{initial}+u_n) \\ &= \frac52(20+20.04)\\ &= 100.1 \end{align}

There is for buying $$5$$ gold.

If I am going to use all my wallet to buy gold, how many gold can I buy?

• You are almost there. Instead of numbers use variables. In your last equation replace un with the expression above. You should get a quadratic equation in $n$. Solve it to find n. May 15, 2022 at 17:27
• Sure, I am going to learn quadratic equation. So, we do not need to state the n, we need to find n value. Anyway, do you have reference regarding this case? May 15, 2022 at 17:56
• @BobbyJulian Don't google it $-$ take some time to try to solve $x^2 - 5x + 6 = 0$ yourself. It won't be easy, but it will be worth it! Once you get the answer, or if you are seriously stuck, you can learn how to solve a generalised quadratic equation here.
– user905694
May 15, 2022 at 18:09
• Sure, I am watching youtube now. May 15, 2022 at 18:12
• @BobbyJulian Once you understand arithmetic progressions, I suggest you try to solve $x^2 - 5x + 6 = 0$ yourself. It will be a fun challenge.
– user905694
May 15, 2022 at 18:14

Let's say $$c$$ is the total cash you pay, $$n$$ is the number of gold, $$p$$ is the price per gold and $$i$$ is the increment. Thus, following the same steps you took (but in general) we get \begin{align} c &= p +(p+i) + (p+2i) + \dots + (p+(n-1)i)\\ & = np +i(1+2+3+\dots+n-1)\\ & = np +i\frac{(n-1)n}{2}\\ & = \frac{n}{2}\left(2p+i(n-1)\right) \end{align} So now we want to solve for $$n$$ given some cash $$c$$ available, we get \begin{align} 2c = n\left(2p+in-i\right) &\color{blue}{\implies}n^2(i) + n\left(2p-\frac{2}{2}i\right) = 2\frac{c}{i} \\ &\color{blue}{\implies}n^2 + 2n\left(\frac{p}{i}-\frac{1}{2}\right) +\color{green}{\left(\frac{p}{i}-\frac{1}{2}\right)^2} = 2\frac{c}{i}+\color{green}{\left(\frac{p}{i}-\frac{1}{2}\right)^2}\\ &\color{blue}{\implies} \left\vert n +\frac{p}{i}-\frac{1}{2} \right\vert^2 = 2\frac{c}{i}+\left(\frac{p}{i}-\frac{1}{2}\right)^2\\ &\color{blue}{\implies} n = \frac{1}{2}- \frac{p}{i}\pm\sqrt{2\frac{c}{i}+\left(\frac{p}{i}-\frac{1}{2}\right)^2}\\ &\color{blue}{\implies} n =\frac{i - 2p\pm\sqrt{8ci+\left(2p-i\right)^2}}{2i} \end{align} So since we're looking or the maximum number $$n$$ possible, we get $$n = \left\lfloor\frac{i - 2p+\sqrt{8ci+\left(i-2p\right)^2}}{2i} \right\rfloor$$ where $$\lfloor \cdot \rfloor$$ indicates the floor function.

In your case, plugging in $$c = 2000$$, $$p=20$$ and $$i=0.01$$ gives you the maximum gold amount $$n$$ you can buy.

• Thank you, the accuracy is fantastic. I am still learning the formula. Anyway, how do you get this =𝑛2(2𝑝+𝑖(𝑛−1)) May 15, 2022 at 20:59
• From the previous equation, we have \begin{align}np+i\frac{(n-1)n}{2} &=\color{blue}{\frac{2}{2}}np+i\frac{(n-1)n}{2} \\&=\color{purple}{\frac{n}{2}}2p + \color{purple}{\frac{n}{2}}i(n-1)\\ & =\color{purple}{\frac{n}{2}}\left( 2p + i(n-1)\right) \end{align} May 15, 2022 at 21:59

As observed correctly, the price follows an arithmetic progression.

Hence, we can equate
$$2000 = S_n \quad (S_n = \text{total price to buy} \; n \; \text{gold})$$ $$\therefore 2000 = \frac{n}{2}(2\cdot20 + 0.01(n-1)) = \frac{n}{2}(0.01n + 39.99)$$ $$\therefore 4000 = 0.01n^2 + 39.99n \iff n^2 + 3999n - 400000 = 0$$

Can you take it from here?

• Hi, I understand the calculation. Anyway, what formula do you use. Anyway, how many gold we can buy. I am going to analyse it. May 15, 2022 at 17:53
• @BobbyJulian The same formula you use $-$ $\text{sum} = \frac{n}{2}(\text{initial} + u_n) = \frac{n}{2}(\text{initial} + (\text{initial} + (n-1)\cdot\text{diff})) = \frac{n}{2}(2\cdot \text{initial} + (n-1)\cdot\text{diff})$
– user905694
May 15, 2022 at 17:55
• I am getting stuck here, will be back after watching arithmetic progression. May 15, 2022 at 18:13
• @BobbyJulian Sure, let me know if you have any questions.
– user905694
May 15, 2022 at 18:13