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Find the 7th term of the G.P \[2,-6,18......\]

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Last updated date: 27th Mar 2024
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MVSAT 2024
Answer
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Hint: To find the 7th term which comes in the given geometric progression using the formula \[{{t}_{nth}}=a({{r}^{n-1}})\].

Complete step-by-step answer:
The given geometric progression given to us ,
\[2,-6,18......\]
The first thing we need to do is to find the first term which is \['a'\] and the common ratio \['r'\] from the given progression \[2,-6,18......\]
\[\begin{gathered}
  & a=2\ \text{ }\!\![\!\!\text{ first term which is 2 here }\!\!]\!\!\text{ } \\
 & r=\dfrac{(n+1)th\text{ term}}{nth\text{ term}}=\dfrac{2nd\text{ term}}{1st\ \text{ term}}=\dfrac{-6}{2}=-3 \\
\end{gathered}\]
The formula to find the \[nth\,\text{ term}\] of a geometric progression is…
\[{{t}_{nth}}=a({{r}^{n-1}})\]
Now, since we have the common ratio , first term, the formula to find the geometric progression and the the term which we need to find then we will use the formula…
\[\begin{gathered}
  & {{t}_{nth}}=a({{r}^{n-1}}) \\
 & {{t}_{7th}}=a({{r}^{n-1}})=2{{(-3)}^{7-1}}=2{{(-3)}^{6}} \\
 & {{t}_{7th}}=2{{(3)}^{6}}\ \text{ }\!\![\!\!\text{ we multiply the negative sign six times (}-\times -\times -\times -\times -\times -=\text{+) }\!\!]\!\!\text{ } \\
 & {{\text{t}}_{7th}}=2(729)=1458 \\
 & \\
\end{gathered}\]
Thus the 7th term is 1458.
The value of common ratio was negative thus whenever there was an even power sign it will change into positive. Thus, leading to all the odd numbered term to be positive because after applying the formula \[{{t}_{nth}}=a({{r}^{n-1}})\] the \[(n-1)term\] factor will change the odd term to an even number thus giving us the positive values in the odd terms.
\[\begin{gathered}
  & {{t}_{3rd}}=a({{r}^{n-1}})=2{{(-3)}^{3-1}}=2{{(-3)}^{2}} \\
 & {{t}_{3rd}}=2{{(-3)}^{2}}=2{{(3)}^{2}}=18 \\
\end{gathered}\]

Note: If the given data was as such where instead of using the geometric progression we would have been given the sum of the geometric progression we would have used the formula \[{{\text{t}}_{n}}={{s}_{n}}-{{s}_{(n-1)}}\].
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