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Question

The numbers 1,4,16 can be three terms (not necessarily consecutive) of

A
no AP
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B
only one GP
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C
infinite number of APs
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D
infinite number of GPs
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Solution

The correct options are
C infinite number of APs
D infinite number of GPs
4=1+(n1)d, 16=1+(m1)d153=m1n1
n11=m15=p= positive integer

n=p+1, m=5p+1.
So, n, m have infinite pairs of values.

Also, 4=1rn, 16=1rm
r2n=rm
m=2n
m2=n1=q= positive integer
So, m, n have infinite pairs of values.

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