Lesson 26.Splan to in you plan to invest 200 intermore than 510.00 annually at the end of the year. How many years until the account intereste than 10,000? 20,000 at the start of each year into an account that averages 3% that at the end of year 1 the amount in the the amount of example, that original 200 200(1.03)^2+200(1.03)=418.18 in the account is 418.18 of year 200 years after the first deposit means adding up deposited has been compounded 0 times pounded a total of 29 times while up suring out how much is in the the account is s=206((1-1.03^n))/((1-1.03)) Now we have a simpler expression 200(1.03)^29+200(1.03)^28+ldots +200(1.03)+200 2000enricsequence,s=a((1-r^n))/((1-r)), to find th We can use the formula for the sum of a=200 and increases at a rate of r=1.03 each year. After n years, the total s in . It turns out that when n=31 n=47 it has about 20,682 in it. the account has about 10,301 in it and when

## Step 1: Calculate the sum after n years<br />### The problem asks how many years it takes for an account to reach a certain amount given an initial investment of $200 at the beginning of each year with a 3% annual interest rate compounded annually. We can use the formula for the sum of a finite geometric series to calculate the total amount in the account after n years: $S_n = P \cdot \frac{(1+r)^{n} - 1}{r}$, where $S_n$ is the sum after n years, P is the annual investment ($200), and r is the interest rate (0.03).<br /><br />## Step 2: Solve for n when the sum reaches $10,000<br />### Substitute $S_n = 10000$, $P = 200$, and $r = 0.03$ into the formula: $10000 = 200 \cdot \frac{(1.03)^{n} - 1}{0.03}$. Simplify and solve for n: $1.5 = (1.03)^n - 1$, $(1.03)^n = 2.5$. Taking the logarithm of both sides: $n \cdot \log(1.03) = \log(2.5)$, $n = \frac{\log(2.5)}{\log(1.03)} \approx 30.9$. Since n represents the number of years, we round up to the nearest whole number, so it takes approximately 31 years.<br /><br />## Step 3: Solve for n when the sum reaches $20,000<br />### Substitute $S_n = 20000$, $P = 200$, and $r = 0.03$ into the formula: $20000 = 200 \cdot \frac{(1.03)^{n} - 1}{0.03}$. Simplify and solve for n: $3 = (1.03)^n - 1$, $(1.03)^n = 4$. Taking the logarithm of both sides: $n \cdot \log(1.03) = \log(4)$, $n = \frac{\log(4)}{\log(1.03)} \approx 47.5$. Since n represents the number of years, we round up to the nearest whole number, so it takes approximately 48 years.