Official Solution:

The sequence \(a_1\), \(a_2\), \(a_3\), ... , \(a_n\), ... is such that \(a_n=a_{n-1}+a_{n-2}\) for all \(n \ge 3\). If \(a_1 = 3\) and \(a_5 = 6\), what is the value of \(a_2\) ?

A. -3
B. -1
C. 0
D. 2
E. 3


Since given that \(a_n=a_{n-1}+a_{n-2}\), then:

\(a_5=a_4+a_3=(a_3+a_2)+(a_2+a_1)=((a_2+a_1)+a_2)+(a_2+a_1)=3*a_2+2*a_1\).

So, we have that \(a_5=3*a_2+2*a_1\), or \(6=3*a_2+2*3\), which gives \(a_2=0\).


Answer: C
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Nth term = (N-1)th term + (n-2)th term

Just substitute the values:

a1=3
a2=x
a3=a2+a1 =x+3
a4=a3+a2 =x+3+x =2x+3
a5=a4+a3 =2x+3+x+3 = 3x+6

now a5=6, so 3x+6=6 ~ 3x=0 ~ x=0.
Hence, a2=x=0.

This is lengthy process though!