1. **Problem Statement:** A person deposits money for three consecutive months forming a geometric progression (GP) with total sum 65. If the first and third amounts are multiplied by 3 and the second by 5, these products form an arithmetic progression (AP). Find the first and second deposit amounts. 2. **Define variables:** Let the first deposit be $a$, the common ratio be $r$. Then the three deposits are $a$, $ar$, and $ar^2$. 3. **Sum of deposits:** Since they form a GP, the sum is: $$a + ar + ar^2 = 65$$ Simplify: $$a(1 + r + r^2) = 65$$ 4. **Condition for AP:** The products are $3a$, $5ar$, and $3ar^2$. These form an AP, so the middle term is the average of the extremes: $$2 \times 5ar = 3a + 3ar^2$$ Simplify: $$10ar = 3a + 3ar^2$$ Divide both sides by $a$ (assuming $a \neq 0$): $$10r = 3 + 3r^2$$ Rearranged: $$3r^2 - 10r + 3 = 0$$ 5. **Solve quadratic for $r$:** Using quadratic formula: $$r = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 3 \times 3}}{2 \times 3} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm 8}{6}$$ Two solutions: - $$r = \frac{10 + 8}{6} = 3$$ - $$r = \frac{10 - 8}{6} = \frac{1}{3}$$ 6. **Find $a$ for each $r$:** From step 3: $$a(1 + r + r^2) = 65$$ - For $r=3$: $$a(1 + 3 + 9) = 65 \Rightarrow a \times 13 = 65 \Rightarrow a = 5$$ - For $r=\frac{1}{3}$: $$a\left(1 + \frac{1}{3} + \frac{1}{9}\right) = 65 \Rightarrow a \times \frac{13}{9} = 65 \Rightarrow a = 65 \times \frac{9}{13} = 45$$ 7. **Check deposits:** - For $r=3$, deposits are $5$, $15$, $45$. - For $r=\frac{1}{3}$, deposits are $45$, $15$, $5$. Both satisfy the problem conditions. **Final answer:** The first deposit is either $5$ or $45$, and the second deposit is $15$ in both cases.