Acarteus
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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.

Verified Answer :

caylus

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

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SociometricStar

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

\(\frac{x}{18}=\frac{8}{x}\\\\x^2=144\\x=\pm12\)

So, there are two possible sequences

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

and

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

Thus, the common ratios are

\(r=\frac{-12}{18},\frac{12}{18}\\\\r=\frac{-2}{3},\frac{2}{3}\)

The sum of the infinite geometric for \(r=\frac{-2}{3}\)

\(S=\frac{a}{1-r}\\\\\frac{18}{1-\frac{-2}{3}}\\\\S=\frac{18}{\frac{5}{3}}\\\\S=\frac{54}{5}\)

The sum of the infinite geometric for \(r=\frac{2}{3}\)

\(S=\frac{a}{1-r}\\\\\frac{18}{1-\frac{2}{3}}\\\\S=\frac{18}{\frac{1}{3}}\\\\S=54\)

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