Question

I need help on these other 2 sequences.
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Answer 1

591.09375

1/4

Explanation:

"Find the sum of the first 8 terms of the geometric sequence that begins: 12, 18, 27,..."

The sum of the first n terms of a geometric sequence is:

S = a (1 − r^n) / (1 − r)

where a is the first term and r is the common ratio.

Here, the first term is 12, so a = 12. The common ratio is 18/12 = 1.5. So the sum of the first 8 terms is:

S = 12 (1 − 1.5^8) / (1 − 1.5)

S = 591.09375

"Find the sum of the infinite geometric sequence that begins 1/12, 1/18, 1/27,..."

The sum of an infinite geometric sequence is:

S = a / (1 − r)

Here, the first term is 1/12, so a = 1/12. The common ratio is (1/18) / (1/12) = 2/3. So the infinite sum is:

S = 1/12 / (1 − 2/3)

S = 1/4

Answer 2

bear in mind that one can always get the common ratio by simply dividing any of the terms by the one before it namely, say 18÷12 or 27÷18.

`\bf 12~~,~~\stackrel{12\cdot (3)/(2)}{18}~~,~~\stackrel{18\cdot (3)/(2)}{27}\qquad \qquad \stackrel{\textit{common ratio}}{r = \cfrac{3}{2}} \\\\[-0.35em] ~\dotfill`

`\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ \displaystyle S_n=\sum\limits_(i=1)^(n)\ a_1\cdot r^(i-1)\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ r=\textit{common ratio}\\ \cline{1-1} a_1=12\\ r=(3)/(2)\\ n=8 \end{cases}`

`\bf S_8=12\left( \cfrac{1-\left( (3)/(2) \right)^8}{1-(3)/(2)} \right)\implies S_8=12\cdot \cfrac{(-6305)/(256)}{-(1)/(2)}\implies S_8=12\cdot \cfrac{6305}{128} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill S_8=\cfrac{18915}{32}~\hfill`