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

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

⇒So, From  , 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 .

⇒When, n=2

Number of terms which are Positive  Integers =2 which is .

⇒When, n=3

Number of terms which are Positive Integers =3, which is .

..........

So,⇒ when , n=r

Number of terms which are Positive Integers =r, which is .

4

step-by-step explanation:

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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