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# If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) =

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Re: If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = [#permalink]
Given , f(x + 2) = f(x) + f(x + 1)
f(15) = f(13) + f(14)
f(13) + f(14) = 617 —————1)
f(12) + f(13) = f(14) ————-2)
f(11) + f(12) = f(13)————-3)
from eq 1) , 2) & 3)
2f(11) + 3f(12) = 617
Putting f(11) = 91
f(12) = (617 – 2*91)/3 = 145
f(12) = f(11) + f(10)
so f(10) = f(12) – f(11) = 145 – 91 =54
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Re: If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = [#permalink]
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Bunuel wrote:
If $$f(x + 2) = f(x) + f(x + 1)$$ for all positive integers x, and $$f(11) = 91$$, $$f(15) = 617$$, then what is the value of $$f(10)$$ ?

A. 53
B. 54
C. 55
D. 56
E. 57

617=f14+f13=f13+f12+f13=2(f13)+f12=2(f12+f11)+f12
617=3f12+91, f12=435/3=145, f12=f11+f10
145=91+f10, f10=54

ans (B)
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Re: If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = [#permalink]
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Given that $$f(x + 2) = f(x) + f(x + 1)$$ for all positive integers x , and $$f(11) = 91$$, $$f(15) = 617$$ and we need to find the value of $$f(10)$$

$$f(x + 2) = f(x) + f(x + 1)$$ - this means that f(any positive integer) = sum of f of previous two numbers
=> f(15) = f(13) + f(14) ...(1)
f(14) = f(12) + f(13) ...(2)
f(13) = f(11) + f(12) ....(3)
f(12) = f(10) + f(11)

Putting value of f(12) in (3) we get
f(13) = f(11) + f(10) + f(11) = f(10) + 2*f(11)

Putting value of f(13) and f(12) in (2) we get
f(14) = f(10) + f(11) + f(10) + 2*f(11) = 2*f(10) + 3*f(11)

Putting value of f(14) and f(13) in (1) we get
f(15) = f(10) + 2*f(11) + 2*f(10) + 3*f(11) = 3*f(10) + 5*f(11)
=> f(10) = $$\frac{f(15) - 5*f(11) }{3}$$ = $$\frac{617 - 5*91}{3}$$ = $$\frac{617-455}{3}$$ = $$\frac{162}{3}$$ = 54

So, Answer will be B
Hope it helps!

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Re: If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = [#permalink]
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