1. ## Geometric Progression

Q1 a G.P is such that S2 = 60 and S3 = 114 find the first 3 terms

Q2 8000 tons of a particular mineral were mined in each of the first 3 yrs of a mines operation. in the fourth yr the quantity mined was 90% of the third year's output, the fifth yr was 90% of the fourth yr and so on. the mine was closed at the end of the first year in which the amount mined fell below 1900 tons
a) for how many yrs did the mine stay open?
b) what total tonnage of the mineral was obtained from the mine during its lifetime?

*i dont even understand wot the Question is asking

2. Hello, Fibonacci!

1) A G.P is such that $\displaystyle S_2 = 60\text{ and }S_3 = 114$. . Find the first 3 terms
Sum of the first $\displaystyle n$ terms: .$\displaystyle S_n \;=\;a\,\frac{r^n-1}{r-1}$

$\displaystyle S_3=114\quad\Rightarrow\quad a\,\frac{r^3-1}{r-1} \:=\:114\;\;{\color{blue}[1]}$

$\displaystyle S_2 = 60 \quad\Rightarrow\quad a\,\frac{r^2-1}{r-1} \:=\:60\;\;{\color{blue}[2]}$

Divide [1] by [2]: .$\displaystyle \frac{a\,\frac{r^3-1}{r-1}}{a\,\frac{r^2-1}{r-1}} \;=\;\frac{114}{60} \quad\Rightarrow\quad \frac{r^3-1}{r^2-1} \:=\:\frac{19}{10}$

. . $\displaystyle \frac{(r-1)(r^2+r+1)}{(r-1)(r+1)} \:=\:\frac{19}{10} \quad\Rightarrow\quad \frac{r^2+r+1}{r+1} \:=\:\frac{19}{10}$

. . which simplifies to: .$\displaystyle 10r^2 - 9r - 9 \:=\:0 \quad\Rightarrow\quad (3r-3)(5r+3) \:=\:0$

. . Hence: .$\displaystyle r \;=\;\frac{3}{2},\;-\frac{3}{5}$

But since two consecutive terms of the G.P. are positive, $\displaystyle r$ cannot be negative.

. . Therefore: .$\displaystyle \boxed{r \;=\;\frac{3}{2}}$

Substitute into [2]: .$\displaystyle a\,\frac{\left(\frac{3}{2}\right)^2 - 1}{\frac{3}{2}-1} \:=\:60 \quad\Rightarrow\quad\frac{5}{2}a \:=\:60 \quad\Rightarrow\quad\boxed{ a \:=\:24}$

Therefore, the first three terms are: .$\displaystyle 24,\:36,\:54$