F 1-2- f 2-3f n - f n -1- f n -2 for n - 1- then what is f 123 -A- -2B- -1C- 1D- 2

The correct option is C 1f_1=2, f_2=3 f_n+2 = f_n+1 f_n, n ≥ 1 f_3 = f_2 f_1=3 2=1 f_4 = f_3 f_2=1 3= 2 f_5= f_4 f_3 = 2 1= 3 f_6 = f_5 f_4 = 3 + 2 = ...

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Quantitative Aptitude

Functions

f 1=2, f 2=3f...

Question

$f_{1} = 2, f_{2} = 3$
$f_{n} = f_{n + 1} - f_{n + 2}$ for n $\geq$ 1, then what is $f_{123}$ ?

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Solution

The correct option is C 1
$f_{1} = 2, f_{2} = 3 f_{n + 2} = f_{n + 1} - f_{n}, n \geq 1 f_{3} = f_{2} - f_{1} = 3 - 2 = 1 f_{4} = f_{3} - f_{2} = 1 - 3 = - 2 f_{5} = f_{4} - f_{3} = - 2 - 1 = - 3 f_{6} = f_{5} - f_{4} = - 3 + 2 = - 1 f_{7} = f_{6} - f_{5} = - 1 + 3 = 2 f_{8} = f_{7} - f_{6} = 2 + 1 = 3 f_{9} = f_{8} - f_{7} = 3 - 2 = 1$
The values of $f_{n}$ will be repeated after n = 6
$f_{123}$ $\Rightarrow$ $\frac{123}{6}$ $\Rightarrow$ Remainder = 3
$f_{123} = f_{3} = 1$

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f1=2,f2=3 fn=fn+1−fn+2 for n ≥ 1, then what is f123?

The correct option is C 1f1=2,f2=3fn+2=fn+1−fn, n≥1f3=f2−f1=3−2=1f4=f3−f2=1−3=−2f5=f4−f3=−2−1=−3f6=f5−f4=−3+2=−1f7=f6−f5=−1+3=2f8=f7−f6=2+1=3f9=f8−f7=3−2=1 The values of fn will be repeated after n = 6 f123 ⇒ 1236 ⇒ Remainder = 3 f123=f3=1

$f_{1} = 2, f_{2} = 3$
$f_{n} = f_{n + 1} - f_{n + 2}$ for n $\geq$ 1, then what is $f_{123}$ ?