Progression and profit and loss question

Question

A merchant can buy goods at the rate of 20 dollars per good. The particular good is part of an overall collection and the value is linked to the number of items that are already on the market. So, the merchant sells the first good for 2 dollars, second good for 4 dollars, third for 6 dollars... and so on. If he wants to make a profit of at least 40%, what is the minimum number of goods he should sell?

I assume the number of goods to be n. Total cost price becomes 20n. (2+4+6...nth term) is equal to or greater than 1.4*20n 2(1+2+3...nth term) is greater than or equal to 28n. The next step in the explanatory answer says n(n+1) is equal to or greater than 28n.I know how to solve from here but I don't understand where this n(n+1) came from. The answer is 27. Can somebody please tell me?

Answer 1

The sum of the first n integers (the range 1..n) can be computed by n(n+1)/2 (you can check that 1+2+...+10 is equal to 10(10+1)/2.)

In your case, you want twice that sum, i.e. n(n+1).

More details on the formula for the sum of an arithmetic series is available e.g. at Wikipedia: https://en.m.wikipedia.org/wiki/Arithmetic_progression

Answer 2

If he wants to make a profit of at least 40%,

Usually one should specify if $40\%$ is referred to the total cost or the total revenue...in this case it is referred to the total cost but usually in a P&L report the profit percentage is referred to the total revenue.

Given this, and remembering that

$$S=1+2+3+4+5+6+\dots+n=\frac{n(n+1)}{2}$$

the sum of your revenue is

$$2\times S=n(n+1)$$

Thus to solve your problem you can set this inequality

$$\frac{20n}{n(n+1)}\geq \frac{1}{1.4}$$

getting

$$n\geq 27$$