A worker was paid a salary of $8,400 in 1982. Each year, a salary increase of 5% of the previous year's salary was awarded. How much did the worker earn from the beginning of 1982 through the end of 2000? (Round your answer to the nearest cent.)

Hint: Sn= a1(1-rn)/1-r, r≠ 1, where a1 is the first term and r is the common ratio.

A. $56,247.87
B. $256,527.63
C. $183,629.31
D. $211,138.34

Question

A worker was paid a salary of $8,400 in 1982. Each year, a salary increase of 5% of the previous year's salary was awarded. How much did the worker earn from the beginning of 1982 through the end of 2000? (Round your answer to the nearest cent.)

Hint: Sn= a1(1-rn)/1-r, r≠ 1, where a1 is the first term and r is the common ratio.

A. $56,247.87
B. $256,527.63
C. $183,629.31
D. $211,138.34

mathmale · Answer

This surely sounds like compounding (a term that you see more often in finanance).  More specifically:  annual compounding.    But the formula given to you as a hint is for finding the "nth sum of a geometric series."  a1 is the first term, the initial value.  r is the annual interest rate, and n is the number of years.

 Sn= a1(1-rn)/1-r, r≠ 1"  

Key words here:  "geometric series," "compounding," "common ratio."

mathmale · Answer

First of all, what have you already learned about problems of this type in class or online?