Salary If you take a job with a starting salary of 35,000...

Salary If you take a job with a starting salary of $\$ 35,000$ per year and a guaranteed raise of $\$ 1400$ per year, how many years will it be before your aggregate salary is $\$ 280,000 ?$ I Hint: Remember that your aggregate salary after 2 years is $\$ 35,000+(\$ 35,000+\$ 1400) .]$

Key Concepts

Arithmetic Sequence

An arithmetic sequence is a list of numbers where each term after the first is obtained by adding a constant difference. This concept is important in salary progression problems because the salary increases by a fixed amount each year, forming a sequence with a first term representing the starting salary and a common difference representing the annual raise.

Sum of an Arithmetic Series

The sum of an arithmetic series is calculated by adding all the terms of an arithmetic sequence. The formula, which is typically written as S = n/2 * (2a + (n - 1)d) or S = n/2 * (first term + last term), allows one to quickly compute the total earnings over a number of years. This is crucial when determining how many years it will take for the cumulative salary to reach a target amount.

Quadratic Equation

When equating the sum of an arithmetic series to a target aggregate salary, the resulting equation is often a quadratic equation in terms of the number of years. Solving this equation requires techniques such as factoring, the quadratic formula, or completing the square, which ultimately provides the time frame needed to achieve the desired total salary.

Transcript

00:01 So in this problem, we're told that our starting salary is $35 ,000.

00:05 So that's going to be our a -1 term.

00:06 And we're told that each year we're going to get a guaranteed raise of $1 ,400.

00:11 So that's our common difference, $1 ,400.

00:13 We want to know how many years it will be before our aggregate salary is $280 ,000.

00:19 Well, aggregate means the sum of.

00:21 So we want to figure out after how many years, will the sum of each of our year's salaries will equal $280 ,000.

00:28 So we're looking for our n value because we want to know how many years it's going to take.

00:33 So to do this, we're going to do s of n is equal to n divided by 2 times the quantity of 2 times a sub 1 plus the quantity of n minus 1 times d.

00:46 So let's substitute in the values that we know.

00:48 Well, we have s of n, which is 280 ,000 equals n which we don't know, divided by 2 times the quantity of 2 times 35 ,000.

01:01 Plus n minus 1 times d, which is 1 ,400.

01:08 So to solve, what i'm going to do first is multiply both sides of my equation by 2.

01:12 Well, 280 ,000 times 2 is equal to 560 ,000.

01:20 Now, this is going to cancel our 2s, so we're just going to have n.

01:23 And then 2 times 35 ,000 is 70 ,000.

01:29 And then we're going to distribute the 1 ,400.

01:31 So we'll have plus 1 ,400 n minus 1 ,400.

01:37 Well, then we can combine our like terms.

01:40 We have 70 ,000 minus 1 ,400, which is equal to 68 ,600.

01:47 So we're left with 560 ,000 equals n times 68 ,600 plus 1 ,400.