Question

question 17
consider a population that grows according to the recursive rule $p_n = p_{n - 1}+ 120$, with initial population $p_1 = 150$.
then:
$p_1 = $
$p_2 = $
find the function for the population. your formula should involve $n$ (use lowercase $n$)
$p(n) = $
use your explicit formula to find $p(100)$
$p(100) = $
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Explanation:

Step1: Find \( P_1 \)

The problem states that the initial population \( P_1 = 150 \), so we directly use this given value.
\( P_1 = 150 \)

Step2: Find \( P_2 \)

Using the recursive rule \( P_n = P_{n - 1}+ 120 \), for \( n = 2 \), we have \( P_2 = P_{1}+ 120 \). Substitute \( P_1 = 150 \) into the formula.
\( P_2 = 150 + 120 = 270 \)

Step3: Find the explicit formula

The recursive rule \( P_n = P_{n - 1}+ 120 \) represents an arithmetic sequence with first term \( a_1 = P_1 = 150 \) and common difference \( d = 120 \). The formula for the \( n \)-th term of an arithmetic sequence is \( a_n=a_1+(n - 1)d \). Substituting \( a_1 = 150 \) and \( d = 120 \), we get:
\( P(n)=150+(n - 1)\times120 \)
Simplify the formula:
\( P(n)=150 + 120n-120=120n + 30 \)

Step4: Find \( P(100) \)

Substitute \( n = 100 \) into the explicit formula \( P(n)=120n + 30 \).
\( P(100)=120\times100+30=12000 + 30=12030 \)

Answer:

\( P_1=\boldsymbol{150} \)
\( P_2=\boldsymbol{270} \)
\( P(n)=\boldsymbol{120n + 30} \)
\( P(100)=\boldsymbol{12030} \)