## Wednesday, January 2, 2013

### Arithmetic Progression - Salary Increase

Category: Algebra

"Published in Newark, California, USA"

Noel was hired by a local petroleum refinery at a starting salary of ₱ 40,000 a month. If the company gives him a ₱ 30,000 increase in his annual salary every year, how much would he be getting at the start of his 10th year with the company? If he regularly saves 10% of his monthly salary, what will be his total savings at the start of the 10th year?

Solution:

From the analysis of a given word problem, this is an application of arithmetic progression because there's a common difference in the sequence which is ₱ 30,000 per year, the amount of annual salary increase of Noel.

The initial annual salary of Noel = ₱ 40,000 (12) = ₱ 480,000.

First, the nth term of an Arithmetic Progression is give by the formula

where

a1first term of an arithmetic progression

an = last term of an arithmetic progression

n = nth term of an arithmetic progression

d = common difference of an arithmetic progression

In the given word problem,

an = amount of Noel's salary at the end of his 10th year with the company

a1initial salary of Noel which is ₱ 480,000

n = number of years in working with the company which is 10 years

d = amount of annual salary increase of Noel which is ₱ 30,000

Substitute the given items to the above equation in order to get the amount of Noel's salary at the end of his 10th year with the company as follows

Therefore, the amount of Noel's salary at the end of his 10th year with the company is ₱ 750,000. We can calculate his monthly salary at the start of his 10th year with the company as follows

Therefore, his monthly salary at the start of his 10th year with the company is ₱ 62,500.

Next, the Sum of an Arithmetic Progression is given by the formula

where

Sn = sum of the terms of an arithmetic progression from a1 to an

a1 = first term of an arithmetic progression

n = nth term of an arithmetic progression

d = common difference of an arithmetic progression

In the given word problem,

a1 = initial salary of Noel which is ₱ 480,000

S9 = sum of Noel's salary at the start of his 10th year with the company. In this calculation, we will use 9 years because the problem says "at the start of his 10th year".

n = number of years in working with the company which is 10 years. In this calculation, we will use 9 years because the problem says "at the start of his 10th year".

d = amount of annual salary increase of Noel which is ₱ 30,000

Substitute the given items to the above equation in order to get the sum of Noel's salary at the start of his 10th year with the company as follows

Therefore, his total savings at the start of his 10th year with the company is

Savings = 0.10 S9

Savings = 0.10 (₱ 5,400,000)

Savings = ₱ 540,000

Note: The symbol, ₱ is Philippine Pesos.