Find the 10th term of the sequence defined by the given rule. assume that the domain of each function is the set of whole numbers greater than 0. f(1) = 40, f(n) = 2 × f(n − 1) − 60

The 10th term of the sequence defined by the given rule is -10180

Given the following recursive functions

Get f(2)

f(2) = 2f(1) - 60

f(2) = 2(40) - 60

f(2) = 80 -60

f(2) = 20

Get f(3)

f(3) = 2f(2) - 60

f(3) = 2(20) - 60

f(3) = 40 -60

f(3) = -20

Get f(4)

f(4) = 2f(3) - 60

f(4) = 2(-20) - 60

f(4) = -40 -60

f(4) = -100

Get f(5)

f(5) = 2f(4) - 60

f(5) = 2(-100) - 60

f(5) = -200 -60

f(5) = -260

Get f(6)

f(6) = 2f(5) - 60

f(6) = 2(-260) - 60

f(6) = -520 -60

f(6) = -580

Get f(7)

f(7) = 2f(6) - 60

f(7) = 2(-580) - 60

f(7) = -1160 -60

f(7) = -1220

Get f(8)

f(8) = 2f(7) - 60

f(8) = 2(-1220) - 60

f(8) = -2440 -60

f(8) = -2500

Get f(9)

f(9) = 2f(8) - 60

f(9) = 2(-2500) - 60

f(9) = -5000 -60

f(9) = -5060

Get f(10)

f(10) = 2f(9) - 60

f(10) = 2(-5060) - 60

f(10) = -10120 -60

f(10) = -10180

Hence the 10th term of the sequence defined by the given rule is -10180

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f(1)=40

f(2) = 2(40)-60 = 20

f(3) = 2(20)-60 = -20

f(4) = 2(-20)-60= -100

f(5) = 2(-100) - 60 = -260

f(6) = 2(-260) - 60 = -580

We could clearly go on like this and get to f(10) but I suspect we're supposed to find the general form.

Each step doubles and subtracts  so we expect a form

Subtracting,

We suspect

Let's check:

It looks like we have the correct formula

-10180

step-by-step explanation: