Determine all nonnegative integers $r$ such that it is possible for an infinite geometric sequence to contain exactly $r$ terms that are integers. Prove your answer.

Let, the geometric sequence is such that, value of common ratio is less than 1.

The Sequence is

$6^{n-1},6^{n-2},6^{n-3},.....,.......\infinity.$

The Geometric Squence is infinite geometric sequence, as there are uncountable terms in the sequence.

⇒So, From $6^{n-1}$ , to infinity, there will be n terms which will be integers when , n≥1.

⇒Put, n=1,

Number of terms which are Positive Integers =1 which is $6^{n-1}$ .

⇒When, n=2

Number of terms which are Positive Integers =2 which is $6^{n-1},6^{n-2}$ .

⇒When, n=3

Number of terms which are Positive Integers =3, which is $6^{n-1},6^{n-2},6^{n-3}$ .

..........

So,⇒ when , n=r

Number of terms which are Positive Integers =r, which is $6^{n-1},6^{n-2},6^{n-3},6^{n-4},........6^{n-r}$ .

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