The first term of an infinite geometric sequence is 18, while the third term is 8. There are two possible sequences. Find the sum of each sequence.

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Let'q the ratio
[tex] u_{1} =18\\ u_{2} =18*q\\ u_{3} =18*q^2=8\ ==\ \textgreater \ q^2= \frac{4}{9} ==\ \textgreater \ q=\pm\ \frac{2}{3} \\ 1) if \ q= \dfrac{2}{3} \ then \ \sum_{i=0}^{\infty}\ 18* (\frac{2}{3} )^i= 18*\frac{1}{1+ \frac{2}{3} } = \frac{18*3}{5}= 10.8\\\\ 2) if \ q= -\dfrac{2}{3} \ then \ \sum_{i=0}^{\infty}\ 18* (-\frac{2}{3} )^i= 18*\frac{1}{1- \frac{2}{3} } = \frac{18*3}{1}= 54\\\\ [/tex]

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SociometricStar SociometricStar · Answer 2 · 2019-06-25T10:36:42+02:00

Answer:

Step-by-step explanation:

The sequence is geometric with

a = 18

Third term = 8.

Let the second term be x.

The infinite geometric sequence will be in the form

18,x,8,......

For geometric sequence, the ratio of two consecutive terms are equal. Hence, we have

[tex]\frac{x}{18}=\frac{8}{x}\\\\x^2=144\\x=\pm12[/tex]

So, there are two possible sequences

18,-12,8,......

and

18,12,8,......

Thus, the common ratios are

[tex]r=\frac{-12}{18},\frac{12}{18}\\\\r=\frac{-2}{3},\frac{2}{3}[/tex]

The sum of the infinite geometric for [tex]r=\frac{-2}{3}[/tex]

[tex]S=\frac{a}{1-r}\\\\\frac{18}{1-\frac{-2}{3}}\\\\S=\frac{18}{\frac{5}{3}}\\\\S=\frac{54}{5}[/tex]

The sum of the infinite geometric for [tex]r=\frac{2}{3}[/tex]

[tex]S=\frac{a}{1-r}\\\\\frac{18}{1-\frac{2}{3}}\\\\S=\frac{18}{\frac{1}{3}}\\\\S=54[/tex]