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.

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.

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:

$$S_6 = \frac{a(1-r^6)}{1-r} = 182 \quad (**)$$

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:

$$\left(\frac{a(1-r^3)}{1-r}\right)(1+r^3) = 182$$

From equation $(*)$, we know that $\frac{a(1-r^3)}{1-r} = 189$. Substitute this into the equation from Step 3:

$$189(1+r^3) = 182$$

Divide both sides by 189:

$$1+r^3 = \frac{182}{189}$$

Subtract 1 from both sides:

$$r^3 = \frac{182}{189} - 1$$ $$r^3 = \frac{182 - 189}{189}$$ $$r^3 = \frac{-7}{189}$$

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 $r = -\frac{1}{3}$.