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
[tex]f(n) = a 2^n + b[/tex]
[tex]f(1)=40 = 2a + b[/tex]
[tex]f(2) =20 = 4a + b[/tex]
Subtracting,
[tex]-20 = 2a[/tex]
[tex]a = -10[/tex]
[tex]40 = 2a + b = -20 + b[/tex]
[tex]b = 60[/tex]
We suspect
[tex]f(n) =60 -10 (2^n) [/tex]
Let's check:
[tex]f(3) = 60 - 10(8) = -20 \quad\checkmark[/tex]
[tex]f(4) = 60 - 10(16) = -100 \quad\checkmark[/tex]
[tex]f(5) = 60 - 10(32) = -260 \quad\checkmark[/tex]
It looks like we have the correct formula
[tex]f(10) = 60 - 10(2^{10}) = 60 - 10240 = -10180[/tex]
Answer: -10180
The 10th term of the sequence defined by the given rule is -10180
Given the following recursive functions
[tex]f(1) =40\\f(n) = 2 \times f(n -1) - 60[/tex]
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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