Question:
If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals
If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals
Updated On: Jul 29, 2025
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Correct Answer: 54
Solution and Explanation
We are given:
\[ f(x+2) = f(x) + f(x+1) \] \[ f(11) = 91 \]
Step 1: Let \( f(12) = a \)
From the recurrence: \[ f(13) = f(11) + f(12) = 91 + a \] \[ f(14) = f(12) + f(13) = a + (91 + a) = 91 + 2a \] \[ f(15) = f(13) + f(14) = (91 + a) + (91 + 2a) = 182 + 3a \]
Step 2: Use the given \( f(15) = 617 \)
\[ 182 + 3a = 617 \] \[ 3a = 435 \quad \Rightarrow \quad a = 145 \] Thus: \[ f(12) = 145 \]
Step 3: Find \( f(10) \)
From the recurrence: \[ f(12) = f(10) + f(11) \] \[ 145 = f(10) + 91 \] \[ f(10) = 54 \]
✅ Final Answer: \( f(10) = 54 \)
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