Step 1

First we write the formula \(A= \text{Final amount}\ P= \text{principal amount}\ r= \text{rate of interest}\) (in decimals)

\(n=\)number of times principal amount compounded in a year and \(t=\) period of investment

\(A=P\left(1+\frac{r}{n}\right)^{nt}\)

Step 2

Now we put the values \(A= $1200,\)

\(r= 9\%=0.09,\ n =1\) and \(t=10\) years

\(12000=P\left(1+\frac{0.09}{1}\right)^{1\times10}\)

\(12000=P(1.09)^{10}\)

\(12000=P\times2.37\)

\(P=\frac{12000}{2.37}=$5063.29\)

Step 3

Answer: \($5063.29\) to be invested to get the final amount \($12000\) after 10 years compounded annually.

First we write the formula \(A= \text{Final amount}\ P= \text{principal amount}\ r= \text{rate of interest}\) (in decimals)

\(n=\)number of times principal amount compounded in a year and \(t=\) period of investment

\(A=P\left(1+\frac{r}{n}\right)^{nt}\)

Step 2

Now we put the values \(A= $1200,\)

\(r= 9\%=0.09,\ n =1\) and \(t=10\) years

\(12000=P\left(1+\frac{0.09}{1}\right)^{1\times10}\)

\(12000=P(1.09)^{10}\)

\(12000=P\times2.37\)

\(P=\frac{12000}{2.37}=$5063.29\)

Step 3

Answer: \($5063.29\) to be invested to get the final amount \($12000\) after 10 years compounded annually.