Joe cuts pieces of ribbon for a decoration according to a pattern, such that the first piece is 20 cm long, the second piece is 25 cm long and the third piece is 30 cm long. The longest piece of ribbon she requires for the decoration is of length 95 cm.
(i) Show with reasons that a roll of ribbon of length 10 m is sufficient for this decoration.
(ii) For another decoration, the length of the longest piece of ribbon cut as above, is twice the length of the above longest piece. Show by computing, whether two rolls of ribbon of length 10 m each are sufficient for this.
Any Ideas on how to begin?
Hint
You need to show, in the first case, that
$$20+25+30+\cdots + 90 + 95 \le 1000 \;(cm=10 m).$$
This is the sum of consecutive terms of an arithmetic progression. Are you familiar with the sum of terms of an arithmetic progression? See https://en.wikipedia.org/wiki/Arithmetic_progression
Once you have solved this case you can solve the second case by proceeding in a similar way.
Edit
Let's call $S$ the desired sum. Then $$\begin{align} S& = & 20 & + & 25 & + & \cdots & + & 90 & + & 95 \\ S& = & 95 & + & 90 & + & \cdots & + & 25 & + & 20 \\ 2S& = & 115 & + & 115 & + & \cdots & + & 115 & + & 115\end{align}$$ where the second line is the first with orderd reversed and the third line is obtained by suming the first and the second. That is, (note that there are $16$ times $115.$ Compare with the formula $S=\frac{n(a_1+a_n)}{2})$ $$2S=16\cdot 115,$$ or $$S=920 cm < 1000 cm.$$ So, you are done.
For the second case we have to analize if $$20+25+30+\cdots + 185 + 190 \le 2000 \;(cm=20 m).$$ We have that
$$20+25+30+\cdots + 185 + 190 = 3675 cm>20 m).$$ So, we have not ribbon enough in this case.