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Consider a sequence {an} with a1=2 and an=a2n1an2 for all n3, terms of the sequence being distinct. Given that a2 and a5 are positive integers and a5162, then the possible value(s) of a5 can be   


A
162  
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B
64    
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C
32    
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D
2      
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Solution

The correct options are
A 162  
C 32    
a1=2 and anan1=an1an2
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=x48162
x41296
x6
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. (x2, as terms are distinct)

Hence, possible value of a5=448,648
a5=32,162

Mathematics

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