finite
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Each term after second term is sum of two terms immediately preceding terms
=> K3=k2+k1
k4=k3+k2
k4=k2+k1+k2
k4=2k2+k1
//y k5=3k2+2k1
so the terms in the sequence can be generalized in the form of Kn=k2 + (n-3)(k2+k1)
=>K9=k2+6(k1+k2)
and K5=K2+2(k1+k2)
statement 1
K4=11
we know k5 and k4 (this is nothing but two different first degree equations with two unknowns) and hence we can find k1, K2 and thus k9
so statement 1 alone is sufficient
Statement 2
following the same logic as above we can find out k9
so statement 2 alone is sufficient
Hence the answer is D
=> K3=k2+k1
k4=k3+k2
k4=k2+k1+k2
k4=2k2+k1
//y k5=3k2+2k1
so the terms in the sequence can be generalized in the form of Kn=k2 + (n-3)(k2+k1)
=>K9=k2+6(k1+k2)
and K5=K2+2(k1+k2)
statement 1
K4=11
we know k5 and k4 (this is nothing but two different first degree equations with two unknowns) and hence we can find k1, K2 and thus k9
so statement 1 alone is sufficient
Statement 2
following the same logic as above we can find out k9
so statement 2 alone is sufficient
Hence the answer is D
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IMO D
we have K5
1) given K4- we can find K6,K7 and so on
SUFFICIENT
2) given K6- using K5 and K6 we can find K7..and so on
SUFFICIENT
we have K5
1) given K4- we can find K6,K7 and so on
SUFFICIENT
2) given K6- using K5 and K6 we can find K7..and so on
SUFFICIENT
The powers of two are bloody impolite!!