salary increases: A man gets a job with a salary of $30000 a year. He is promised a $2300 raise each subsequent year. Find the total earnings for a 10 year period.

Arithmetic Sequence,:

a sub 1 = $30,000

Use the arithmetic formula
a sub n = a sub 1 + d (n-1) to find how much the man makes in his 10th year, where d = the difference...in this case d = +2300
(When you plug this in it would look like a sub 10 = 30,000 + 2300(10-1) which equals $50700
Then, using this information, plug it into the equation S (sum) sub n equals
n [ (a sub 1 plus a sub n) / 2 ]
(When you plug this in it would look like
S sub 10 = [ (30k + 50700) / 2 ]
This gives you your answer, $403,500

which by the way is what Olaf said :)

Well, let's calculate that with a little bit of math and a whole lot of fun.

In this arithmetic sequence, the first term (a) is $30,000 and the common difference (d) is $2,300. We want to find the total earnings for a 10-year period.

To find the total earnings, we can use the formula for the sum of an arithmetic sequence:

S = (n/2) * (2a + (n-1)d)

Where S is the sum of the terms, n is the number of terms, a is the first term, and d is the common difference.

So, let's plug in the values:

a = $30,000
d = $2,300
n = 10

S = (10/2) * (2 * $30,000 + (10-1) * $2,300)

Now, let's do the math:

S = (10/2) * (2 * $30,000 + 9 * $2,300)
S = 5 * ($60,000 + $20,700)
S = 5 * $80,700
S = $403,500

So, the total earnings for a 10-year period will be $403,500. That's a whole lot of cash!

To find the total earnings for a 10 year period, we need to calculate the salary for each year and then sum them up.

The given information tells us that the man's initial salary is $30,000, and he receives a $2,300 raise each subsequent year.

We can represent this situation using an arithmetic sequence, where the first term (a₁) is $30,000 and the common difference (d) is $2,300.

The formula to find the nth term of an arithmetic sequence is given by:
aₙ = a₁ + (n - 1) * d

Using this formula, we can calculate the salary for each year and then sum them up to find the total earnings for the 10 year period.

Calculating the salary for each year:
a₁ = $30,000 (initial salary)
a₂ = $30,000 + 1 * $2,300
a₃ = $30,000 + 2 * $2,300
a₄ = $30,000 + 3 * $2,300
.
.
.
a₁₀ = $30,000 + 9 * $2,300

To calculate the total earnings, we need to sum up all the salaries from year 1 to year 10:
Total earnings = a₁ + a₂ + a₃ + ... + a₁₀

Now, we just need to substitute the values of a₁, a₂, a₃, ..., a₁₀ into the formula and calculate the sum.

arithmetic sequence

a = 30,000
d = 2,300

n = 10

sum n = 1 to n = 10
of a + d(n-1)

http://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html

= 10/2 (60,000 + 9 * 2,300)

30,000*10 + 2,300*45 = 403,500