### Theory:

In our day to day life, we observe that there are some situations or things that increase in magnitude over a period of time.

- The population of a state/country
- The value of the property
- Investment returns
- The growth of the cell
- The depreciation in the values of machines, vehicles, etc.,

Example:

Similar to the compound interest the above mentions things also will increase over a period. Using particular formulae we can also calculate those things. And we will learn all about this in this chapter.

1. Increase in Growth:

**Population Growth**:

**Type 1)**

**Growth after**\(n\)

**years**:

Let \(P\) be the population of a city or state at the beginning of a certain year and the population grows at a constant rate of \(r \%\) per annum.

Therefore:

Population after \(n\) years \(A\) $=\phantom{\rule{0.147em}{0ex}}P\phantom{\rule{0.147em}{0ex}}{(1+\frac{r}{100})}^{n}$

**Type 2)**

**Growth grows at different rate**:

Let \(P\) be the population of a city or state at the beginning of a certain year and the population grows at a constant rate of \(r_1 \%\) in the first year and, \(r_2 \%\) in the second year.

Population after \(n\) years \(A =\) $P(1+\frac{{r}_{1}}{100})(1+\frac{{r}_{2}}{100})(1+\frac{{r}_{3}}{100}).......(1+\frac{{r}_{n}}{100})$.

**Type 3)**

**Growth decreases**:

Let \(P\) be the population of a city or state at the beginning of a certain year and the population decreases at the rate of \(r \%\) per annum.

Therefore:

Population after \(n\) years $A\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}P\phantom{\rule{0.147em}{0ex}}{(1-\frac{r}{100})}^{n}\phantom{\rule{0.147em}{0ex}}$.

Important!

We can also use the above formulae to the growth of cells in a particular period at a particular rate and the value of land increased or decreased at a particular time at a particular rate.

2. Depreciation:

To find Depreciation if the value of the product or article at certain time \(P\) and rate of depreciation \(r \%\), then we can find the depreciated value at \(n\) years that is:

Depreciated value at \(n\) years \(A\) $=P\phantom{\rule{0.147em}{0ex}}{(1-\frac{r}{100})}^{n}\phantom{\rule{0.147em}{0ex}}$.