Account Growth 8Da9Ca

1. **State the problem:** You have an initial amount $P = 6184.50$ in an account with an annual interest rate $r = 9.06\% = 0.0906$. You plan to deposit $D = 8495.32$ at the beginning of each year. You want to find the time $t$ in months for the account balance to reach $A = 98758.82$. 2. **Formula used:** Since deposits are made at the beginning of each year, this is an annuity due problem with initial principal. The future value $A$ after $n$ years is given by: $$ A = P(1+r)^n + D \times \frac{(1+r)^n - 1}{r} \times (1+r) $$ where $n$ is the number of years. 3. **Rewrite the formula to solve for $n$:** Let $x = (1+r)^n$. Then: $$ A = P x + D \times \frac{x - 1}{r} \times (1+r) $$ Multiply both sides by $r$ to clear denominator: $$ r A = r P x + D (x - 1)(1+r) $$ Expand right side: $$ r A = r P x + D (1+r) x - D (1+r) $$ Group terms with $x$: $$ r A + D (1+r) = x (r P + D (1+r)) $$ Solve for $x$: $$ x = \frac{r A + D (1+r)}{r P + D (1+r)} $$ 4. **Calculate $x$ numerically:** $$ r = 0.0906, \quad P = 6184.50, \quad D = 8495.32, \quad A = 98758.82, \quad 1+r = 1.0906 $$ Calculate numerator: $$ r A + D (1+r) = 0.0906 \times 98758.82 + 8495.32 \times 1.0906 = 8947.77 + 9265.88 = 18213.65 $$ Calculate denominator: $$ r P + D (1+r) = 0.0906 \times 6184.50 + 8495.32 \times 1.0906 = 560.11 + 9265.88 = 9825.99 $$ So: $$ x = \frac{18213.65}{9825.99} = 1.8549 $$ 5. **Solve for $n$:** $$ x = (1+r)^n \Rightarrow n = \frac{\ln(x)}{\ln(1+r)} = \frac{\ln(1.8549)}{\ln(1.0906)} $$ Calculate: $$ \ln(1.8549) = 0.6181, \quad \ln(1.0906) = 0.0867 $$ Therefore: $$ n = \frac{0.6181}{0.0867} = 7.13 \text{ years} $$ 6. **Convert years to months:** $$ \text{months} = 7.13 \times 12 = 85.56 \text{ months} $$ **Final answer:** It will take approximately **85.56 months** to reach your goal.