Originally Posted by
Soroban Hello, Ventus!
Do you know anything about Geometric Series?
The first six terms are: .$\displaystyle a,\; ar,\; ar^2,\; ar^3,\; ar^4,\; ar^5$
We are told: .$\displaystyle \begin{Bmatrix}a + ar + ar^2 \:=\: 21 & \Rightarrow & a(1+r + r^2) \:=\: 21 & [1] \\ \\[-3mm] ar^3 + ar^4 + ar^5 \:=\: 168 & \Rightarrow & ar^3(1+r+r^2) \:=\: 168 & [2] \end{Bmatrix}$
Divide [2] by [1]: .$\displaystyle \frac{ar^3(1+r+r^2)}{a(1+r+r^2)} \:=\:\frac{168}{21}$
. . And we have: .$\displaystyle r^3 \:=\:8 \quad\Rightarrow\quad \boxed{r \:=\:2}$
Substitute into [1]: .$\displaystyle a(1+2+2^2) \:=\:21 \quad\Rightarrow\quad 7a \:=\:21 \quad\Rightarrow\quad \boxed{a \:=\:3}$
Therefore, the first three terms are: .$\displaystyle (a,\:ar,\:ar^2) \;=\;(3,\:6,\:12)$