1. **State the problem:** We have a geometric progression (GP) where the third term is 6 less than the second term, and the second term is 9 less than the first term. We need to find the fourth term and the sum of the first four terms. 2. **Recall the formula for terms in a GP:** The $n$th term of a GP is given by $$a_n = ar^{n-1}$$ where $a$ is the first term and $r$ is the common ratio. 3. **Set up equations from the problem:** - The second term is $ar$. - The third term is $ar^2$. Given: $$ar^2 = ar - 6$$ (third term is 6 less than second term) $$ar = a - 9$$ (second term is 9 less than first term) 4. **Express $a$ from the second equation:** $$ar = a - 9 \implies a - ar = 9 \implies a(1 - r) = 9 \implies a = \frac{9}{1-r}$$ 5. **Substitute $a$ into the first equation:** $$ar^2 = ar - 6$$ Substitute $a$: $$\frac{9}{1-r} r^2 = \frac{9}{1-r} r - 6$$ Multiply both sides by $(1-r)$: $$9r^2 = 9r - 6(1-r)$$ Simplify right side: $$9r^2 = 9r - 6 + 6r$$ $$9r^2 = 15r - 6$$ 6. **Rearrange to form a quadratic equation:** $$9r^2 - 15r + 6 = 0$$ Divide entire equation by 3: $$3r^2 - 5r + 2 = 0$$ 7. **Solve quadratic equation:** Using quadratic formula: $$r = \frac{5 \pm \sqrt{(-5)^2 - 4 \times 3 \times 2}}{2 \times 3} = \frac{5 \pm \sqrt{25 - 24}}{6} = \frac{5 \pm 1}{6}$$ So, $$r_1 = \frac{5+1}{6} = 1$$ $$r_2 = \frac{5-1}{6} = \frac{4}{6} = \frac{2}{3}$$ 8. **Check for valid $r$:** If $r=1$, then from step 4, denominator $1-r=0$ which is undefined. So discard $r=1$. Thus, $r = \frac{2}{3}$. 9. **Find $a$ using $r=\frac{2}{3}$:** $$a = \frac{9}{1 - \frac{2}{3}} = \frac{9}{\frac{1}{3}} = 27$$ 10. **Find the fourth term:** $$a_4 = ar^3 = 27 \times \left(\frac{2}{3}\right)^3 = 27 \times \frac{8}{27} = 8$$ 11. **Find the sum of the first four terms:** Sum of first $n$ terms of GP: $$S_n = a \frac{1 - r^n}{1 - r}$$ For $n=4$: $$S_4 = 27 \times \frac{1 - \left(\frac{2}{3}\right)^4}{1 - \frac{2}{3}} = 27 \times \frac{1 - \frac{16}{81}}{\frac{1}{3}} = 27 \times \frac{\frac{65}{81}}{\frac{1}{3}} = 27 \times \frac{65}{81} \times 3 = 27 \times \frac{195}{81}$$ Simplify: $$\frac{195}{81} = \frac{65}{27}$$ So, $$S_4 = 27 \times \frac{65}{27} = 65$$ **Final answers:** - Fourth term = 8 - Sum of first four terms = 65