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Sagot :
Answer:
$574
Step-by-step explanation:
[tex]\begin{aligned}\text { week } 1=w_{1} &=300 \\w_{2} &=300+10=310 \\w_{3} &=310+10=320=300+2 \times(10)\end{aligned}[/tex]
It’s obvious here that we add 10 each time to get the next week’s payment so we Are dealing an arithmetic sequence where its first term is 300 and its
Common difference is 10
Then In the twentieth week the payment will be :
[tex]\begin{aligned}W_{20} &=w_{1}+(20-1) \times(10) \\&=300+19 \times(10) \\&=490\end{aligned}[/tex]
Now we apply the formula for the sum of Consecutives terms of an arithmetic sequence :
[tex]\begin{aligned}S_{1} &=20 \times\left(\frac{w_{1}+w_{20}}{2}\right) \\&=20 \times\left(\frac{300+490}{2}\right) \\&=7900\end{aligned}[/tex]
Now let move on to Job B:
[tex]\\\\\begin{aligned}\text { month1 }=& m_{1}=1200 \\m_{2} &=1200+1200 \times \frac{10}{100}=1200 \times(1,1)=1320 \\m_{3} &=m_{2} \times(1,1)=\left[m_{1} \times(1,1)\right] \times(1,1)=m_{1} \times(1,1)^{2} \\m_{4} &=m_{1} \times(1,1)^{3} \\m_{5} &=m_{1} \times(1,1)^{4}\end{aligned}[/tex]
It’s obvious here that we multiply by (1.1) each time to get the next month’s payment so we Are dealing a geometric sequence where its first term is 1200 and its Common ratio is 1.1
Now we apply the formula for the sum of Consecutives terms of an geometric sequence :
[tex]\begin{array}{l}S_{2}=m_{1} \times\left(\frac{1-1,1^{5}}{1-1,1}\right) \\\approx 7326\end{array}[/tex]
Finally we can say confidently say that he would earn more :
7900 - 7326 = $574.
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