Question

Divya deposited Rs. 1000 at compound interest at the rate of 10% per annum. The amount at the end of the first year, second year, third year…. And so on.

Answer 1

Hint: To calculate annual compound interest, multiply the original amount of your investment or loan, or principal, by the annual interest rate. Add the amount to the principle, then multiply by interest rate again to get second year`s compounding interest. So, use this concept to reach the solution of the given problem.

Complete step by step answer:
We know that the compound interest is given by the formula \[A = P{\left( {1 + \dfrac{r}{{n \times 100}}} \right)^{nt}}\] where \[A\] is the final amount, \[P\] is the principle amount, \[t\] is the number of times interest applied per time period, \[n\] is the number of times interest applied per time period and \[r\] is the rate on interest per annum.
For the first year given that
Principal amount \[P = 1000\]
Rate of interest \[r = 10\% \]
Time period in years \[t = 1\]
Number of times interest applied \[n = 1\]
So, the final amount for the first year is
\[
   \Rightarrow {A_1} = 1000{\left( {1 + \dfrac{{10}}{{100 \times 1}}} \right)^{1 \times 1}} \\
   \Rightarrow {A_1} = 1000\left( {\dfrac{{11}}{{10}}} \right) \\
  \therefore {A_1} = 1100 \\
\]
For the second year we have
Principal amount \[P = 1000\]
Rate of interest \[r = 10\% \]
Time period in years \[t = 2\]
Number of times interest applied \[n = 1\]
So, the final amount for the second year is
\[
   \Rightarrow {A_2} = 1000{\left( {1 + \dfrac{{10}}{{100 \times 1}}} \right)^{1 \times 2}} \\
   \Rightarrow {A_2} = 1000{\left( {\dfrac{{11}}{{10}}} \right)^2} \\
  \therefore {A_2} = 10 \times 121 = 1210 \\
\]
For the third year we have
Principal amount \[P = 1000\]
Rate of interest \[r = 10\% \]
Time period in years \[t = 3\]
Number of times interest applied \[n = 1\]
So, the final amount for the third year is
\[
   \Rightarrow {A_3} = 1000{\left( {1 + \dfrac{{10}}{{100 \times 1}}} \right)^{1 \times 2}} \\
   \Rightarrow {A_3} = 1000{\left( {\dfrac{{11}}{{10}}} \right)^3} \\
  \therefore {A_3} = 1000 \times 1.331 = 1331 \\
\]
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For \[{n^{th}}\] year we have
For the third year we have
Principal amount \[P = 1000\]
Rate of interest \[r = 10\% \]
Time period in years \[t = n\]
Number of times interest applied \[n = 1\]
So, the final amount for the \[{n^{th}}\] year is
\[
   \Rightarrow {A_n} = 1000{\left( {1 + \dfrac{{10}}{{100 \times 1}}} \right)^{1 \times n}} \\
   \Rightarrow {A_n} = 1000{\left( {\dfrac{{11}}{{10}}} \right)^n} \\
  \therefore {A_n} = 1000 \times {\left( {1.1} \right)^n} \\
\]

Note: The compound interest is given by the formula \[A = P{\left( {1 + \dfrac{r}{{n \times 100}}} \right)^{nt}}\] where \[A\] is the final amount, \[P\] is the principle amount, \[t\] is the number of times interest applied per time period, \[n\] is the number of times interest applied per time period and \[r\] is the rate on interest per annum.