If the sum of the first three terms of a GP is given by 189 and the sum of the first six terms is given by 182, show that the common ratio is given as r = -1/3.

Step 1: Write the formulas for the sum of the first $n$ terms of a geometric progression (GP). The sum of the first $n$ terms of a GP is given by the formula: $S_n = \frac{a(1-r^n)}{1-r}$ where $a$ is the first term and $r$ is the common ratio.

Step 2: Set up equations based on the given information. We are given that the sum of the first three terms ($S_3$) is 189: $S_3 = \frac{a(1-r^3)}{1-r} = 189 \quad (*)$ We are also given that the sum of the first six terms ($S_6$) is 182: $S_6 = \frac{a(1-r^6)}{1-r} = 182 \quad (**)$

Step 3: Rewrite the expression for $S_6$ using the difference of squares identity. We know that $1-r^6 = (1-r^3)(1+r^3)$. Substitute this into the equation for $S_6$: $S_6 = \frac{a(1-r^3)(1+r^3)}{1-r} = 182$ This can be rearranged as: $\left(\frac{a(1-r^3)}{1-r}\right)(1+r^3) = 182$

Step 4: Substitute the value of $S_3$ into the rearranged equation for $S_6$. From equation $(*)$, we know that $\frac{a(1-r^3)}{1-r} = 189$. Substitute this into the equation from Step 3: $189(1+r^3) = 182$

Step 5: Solve for $r^3$. Divide both sides by 189: $1+r^3 = \frac{182}{189}$ Subtract 1 from both sides: $r^3 = \frac{182}{189} - 1$ $r^3 = \frac{182 - 189}{189}$ $r^3 = \frac{-7}{189}$

Step 6: Simplify the fraction and solve for $r$. Both the numerator and the denominator are divisible by 7: $r^3 = -\frac{7 \div 7}{189 \div 7}$ $r^3 = -\frac{1}{27}$ Take the cube root of both sides to find $r$: $r = \sqrt[3]{-\frac{1}{27}}$ $r = -\frac{1}{3}$

The common ratio is indeed $r = -\frac{1}{3}$.

The final answer is $\boxed{r = -\frac{1}{3}}$.