an employee is paid $3600 for the first month. she is paid an additional 1% at the end of the second month, which means a total of $3636. she continues to receive a 1% raise each month for a full year. what is her total salary at the end of the year?
We have a GS with a = 3600 and r = 1.01 , n = 12
sum(12) = a(r^12 - 1)/(r-1)
= 3600(1.01^12 - 1)/.01
= 45,657.01
The answer in my textbook says its $45,648.
To find the total salary at the end of the year, we need to calculate each month's salary increment and sum them up.
Let's break down the problem step by step:
1. First, we know that the employee's salary in the first month is $3600.
2. In the second month, the employee receives an additional 1%. To calculate the increment, we can subtract the original salary from the total salary at the end of the second month: $3636 - $3600 = $36. This represents a 1% increase.
3. To determine the employee's salary for each subsequent month, we will increase the previous month's salary by 1%.
4. Since the employee receives a 1% raise each month for a full year (12 months), we can calculate the salary increments and sum them up.
Let's calculate the salary for each month:
Month 1: $3600 (given)
Month 2: $3600 + $36 = $3636
Month 3: $3636 + ($3636 * 0.01) = $3672.36
Month 4: $3672.36 + ($3672.36 * 0.01) = $3709.08
...
Month 12: Calculate the salary using the same formula as above.
Now, we need to sum up all 12 monthly salaries to find the total salary at the end of the year.