# In Youngtown, the population reached 4420 people in 1993. In 1998, there were 5600 residents. What was the population in the year 2000?

Jun 19, 2018

Population in $2000$ was $6156$

#### Explanation:

Assuming population growth is an exponential function.

Initial population in 1993 is ${P}_{0} = 4420$

The growth function is P_t=P_0*e^(kt) ; k and t  are growth

rate and time in years.

Population in 1998 is P_5=5600 ; t= 5

$5600 = 4420 \cdot {e}^{5 k} \mathmr{and} {e}^{5 k} = \frac{5600}{4420}$

Taking natural log on both sides we get,

5*k* ln e= ln(280/221)~~ 0.2366 ; (ln e=1) or

$k = \approx \frac{0.2366}{5} \approx \approx 0.0473$

The growth function is ${P}_{t} = 4420 \cdot {e}^{0.0473 t}$

Population in 2000 = ? ; t=2000-1993-7 years

${P}_{t} = 4420 \cdot {e}^{0.0473 \cdot 7} \approx 6155.94$

Population in $2000$ was $6156$ [Ans]

Jun 19, 2018

The population in $2000$ would be $6072$

#### Explanation:

Using Arithmetic Progression..

Let;

First Term ${T}_{1} = a = 4420$

$1993 , \textcolor{w h i t e}{x} 1994 , \textcolor{w h i t e}{x} 1995 , \textcolor{w h i t e}{x} 1996 , \textcolor{w h i t e}{x} 1997 , \textcolor{w h i t e}{x} 1998 , \textcolor{w h i t e}{x} 1999 , \textcolor{w h i t e}{x} 2000$

$\textcolor{w h i t e}{x} \downarrow \textcolor{w h i t e}{\times x} \downarrow \textcolor{w h i t e}{\times x} \downarrow \textcolor{w h i t e}{\times x} \downarrow \textcolor{w h i t e}{\times x} \downarrow \textcolor{w h i t e}{\times x} \downarrow \textcolor{w h i t e}{\times x} \downarrow \textcolor{w h i t e}{\times x} \downarrow$

$\textcolor{w h i t e}{x} {T}_{1} , \textcolor{w h i t e}{\times x} {T}_{2} , \textcolor{w h i t e}{\times} {T}_{3} , \textcolor{w h i t e}{\times x} {T}_{4} , \textcolor{w h i t e}{\times} {T}_{5} , \textcolor{w h i t e}{x , x} {T}_{6} , \textcolor{w h i t e}{x , x} {T}_{7} , \textcolor{w h i t e}{x , x} {T}_{8}$

Therefore;

${T}_{6} \to 1998$

Hence the Sixth Term is $5600$

Now to get the common difference..

Recall;

${T}_{6} = a + 5 d = 5600$

$4420 + 5 d = 5600$

$5 d = 5600 - 4420$

$5 d = 1180$

$d = \frac{1180}{5}$

$d = 236$

Hence;

The common difference is $236$

From the above structure..

${T}_{8} \to 2000$

Hence, the Eight Term is $2000$

Recall;

${T}_{8} = a + 7 d$

${T}_{8} = 4420 + 7 \left(236\right)$

${T}_{8} = 4420 + 1652$

${T}_{8} = 6072$

Therefore, the population in $2000$ would be $6072$