Charlessamparangidn
Answered

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5, 15, 45,.. geometric series of the sum S12​

Sagot :

KAntoineDoixidn

✏️GEOMETRIC SERIES

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[tex] \underline{\mathbb{PROBLEM}:} [/tex]

  • 5, 15, 45,.. geometric series of the sum [tex] \sf S_{12} [/tex]

[tex]\red{••••••••••••••••••••••••••••••••••••••••••••••••••} [/tex]

[tex] \underline{\mathbb{ANSWER}:} [/tex]

[tex] \qquad\Large » \tt\: \green{S_{12} = 1,\!328,\!600 } [/tex]

[tex]\red{••••••••••••••••••••••••••••••••••••••••••••••••••} [/tex]

[tex] \underline{\mathbb{SOLUTION}:} [/tex]

» Determine the common ratio.

[tex] \begin{align} & \bold{Formula:} \\ & \boxed{r = \frac{a_n}{a_{n-1}}} \end{align} [/tex]

  • [tex] r = \frac{a_2}{a_1} = \frac{15}{5} = 3 \\ [/tex]

  • [tex] r = \frac{a_3}{a_2} = \frac{45}{15} = 3 \\ [/tex]

» Find the sum of the first 12 terms of the sequence.

[tex] \begin{align} & \bold{Formula:} \\ & \boxed{S_n = \frac{a_1(1-r^n)}{1-r}} \end{align} [/tex]

  • [tex] S_{12} = \frac{5(1 - 3^{12})}{1 - 3} \\ [/tex]

  • [tex] S_{12} = \frac{5(1 - 531,\!441)}{1 - 3} \\ [/tex]

  • [tex] S_{12} = \frac{5(\text-531,\!440)}{\text-2} \\ [/tex]

  • [tex] S_{12} = \frac{\text-2,\!657,\!200}{\text-2} \\ [/tex]

  • [tex] S_{12} = 1,\!328,\!600 [/tex]

[tex] \therefore [/tex] The sum of the first 12 terms of the geometric sequence is 1,328,600.

[tex]\red{••••••••••••••••••••••••••••••••••••••••••••••••••} [/tex]

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