Question

# Consider a sequence {an} with a1=2 and an=a2n−1an−2 for all n≥3, terms of the sequence being distinct. Given that a2 and a5 are positive integers and a5≤162, then the possible value(s) of a5 can be

A
162
B
64
C
32
D
2

Solution

## The correct options are A 162   C 32    a1=2 and anan−1=an−1an−2 Hence, a1,a2,a3,a4,a5,... are in G.P. Let a2=x. Then for n=3, we have ⇒a3a2=a2a1 ⇒a22=a1a3 ⇒a3=x22  (∵a1=2)  i.e., the sequence is 2,x,x22,x34,x48,... with common ratio, r=x2  Given, a5=x48≤162 ⇒x4≤1296 ⇒x≤6 Also, x and x48 are integers. So, if x is even, then only x48 will be an integer. Hence, possible values of x are 4 and 6. (x≠2, as terms are distinct) Hence, possible value of a5=448,648 ⇒a5=32,162Mathematics

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