What is the sum of the geometric sequence –3, 21, –147, … if there are 6 terms?

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What is the sum of the geometric sequence –3, 21, –147, … if there are 6 terms?

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Answer 1 · 2016-01-25T17:39:48

$\frac{a_{2}}{a_{1}} = \frac{21}{- 3} = - 7$ $a_2/a_1=21/-3=-7$

$\frac{a_{3}}{a_{2}} = - \frac{147}{21} = - 7$ $a_3/a_2=-147/21=-7$

$\Rightarrow$ $implies$ common ratio $= r = - 7$ $=r=-7$ and $a_{1} = - 3$ $a_1=-3$

Sum of geometric series is given by

$S u m = \frac{a_{1} (1 - r^{n})}{1 - r}$ $Sum=(a_1(1-r^n))/(1-r)$

Where $n$ $n$ is number of terms, $a_{1}$ $a_1$ is the first term, $r$ $r$ is the common ratio.

Here $a_{1} = - 3$ $a_1=-3$ , $n = 6$ $n=6$ and $r = - 7$ $r=-7$

$\Rightarrow S u m = \frac{- 3 (1 - {(- 7)}^{6})}{1 - (- 7)} = \frac{- 3 (1 - 117649)}{1 + 7} = \frac{- 3 (- 117648)}{8} = \frac{352944}{8} = 44118$ $implies Sum=(-3(1-(-7)^6))/(1-(-7))=(-3(1-117649))/(1+7)=(-3(-117648))/8=352944/8=44118$

Hence, the sum is $44118$ $44118$ .

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What is the sum of the geometric sequence –3, 21, –147, … if there are 6 terms?

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