Question #8781

Suppose an employee recieves a $2000 raise each year and the sequence a^n models the employee's salary after n years. Is this sequence arithmetic, geometric, or neither? Explain.

Expert's answer

Let C be the initial salary of an employee. Then after n years his salary will be S = C + 2000*n. From the other hand, S = a^n. So, we've got an equation:

C + 2000*n = a^n, n≥1.

Let take n = 1:

C + 2000 = a ==> C = 2000 + a.

Let's take n = 2:

C + 2000*2 = a²,

or

2000 + a + 4000 = a²,

or

a² - a - 6000 = 0& ==>& a = 0.5 ± √6000.25,

or, taking into account that a>0, a = 0.5 + √6000.25.

Let's take n = 3:

C + 2000*3 = 2000 + a + 6000 = 0.5 + √6000.25 + 8000 ≈ 8077.9613,

a³ = (0.5 + √6000.25)³ = 473845.6452;

We see that 8077.9613 < 473845.6452. So, it is impossible to built such a sequence.

C + 2000*n = a^n, n≥1.

Let take n = 1:

C + 2000 = a ==> C = 2000 + a.

Let's take n = 2:

C + 2000*2 = a²,

or

2000 + a + 4000 = a²,

or

a² - a - 6000 = 0& ==>& a = 0.5 ± √6000.25,

or, taking into account that a>0, a = 0.5 + √6000.25.

Let's take n = 3:

C + 2000*3 = 2000 + a + 6000 = 0.5 + √6000.25 + 8000 ≈ 8077.9613,

a³ = (0.5 + √6000.25)³ = 473845.6452;

We see that 8077.9613 < 473845.6452. So, it is impossible to built such a sequence.

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