How do you find the missing terms of the geometric sequence: 3,--,--,--,768?
Answer
587.7k+ views
Hint: This type of question is based on the concept of geometric series. We can solve this question with the help of the expansion of geometric series, that is, \[a,ar,a{{r}^{2}},a{{r}^{3}},,,,,a{{r}^{n-1}},a{{r}^{n}}\]. And from the question, we get, a=3. From the given question, we get to know that the fifth term in the series is 768. Equate the fourth term with \[a{{r}^{4}}\] Then find the value of r and then find the missing terms.
Complete step by step answer:
It is given, the geometric series is 3,-,--,--,768. And we have been asked to find the missing series.
We know that the geometric series is \[a,ar,a{{r}^{2}},a{{r}^{3}},,,,,a{{r}^{n-1}},a{{r}^{n}}\].
We have been given that, the fifth term is 768.
Comparing with the fourth term of geometric series, we get,
\[a{{r}^{4}}=768\] ………………………(1)
From the question, the first term is 3. That means a=3.
Now substitute ‘a’ in equation (1).
\[3{{r}^{4}}=768\] ……………………….(2)
Divide equation (2) by 3. We get,
\[\dfrac{3{{r}^{4}}}{3}=\dfrac{768}{3}\]
\[\Rightarrow {{r}^{4}}=256\]
We know that \[{{4}^{4}}=256\].
We get, \[{{r}^{4}}={{4}^{4}}\].
Taking cube root on both the sides,
\[\sqrt[4]{{{r}^{4}}}=\sqrt[4]{{{4}^{4}}}\]
Since \[\sqrt[4]{{{x}^{4}}}=x\], we get,
\[\therefore r=4\]
Therefore, we obtain the three missing terms as
The first term is
\[ar=3\left( 4 \right)\]
\[\therefore ar=12\].
The second term is
\[a{{r}^{2}}=3{{\left( 4 \right)}^{2}}\]
\[\Rightarrow a{{r}^{2}}=3\left( 16 \right)\]
\[\therefore a{{r}^{2}}=48\].
The third term is
\[a{{r}^{3}}=3{{\left( 4 \right)}^{3}}\]
\[\Rightarrow a{{r}^{3}}=3\left( 64 \right)\]
\[\therefore a{{r}^{3}}=192\]
The three missing terms are 12,48,192.
Hence, the complete geometric series is 3,12,48,192,768.
Note: Whenever we get this type of problem, we have to compare the given series with a geometric series. Arithmetic series should not be used for this type of question. Also, we should avoid calculation mistakes to obtain accurate answers. Avoid mistakes based on sign convention if any. Don’t get confused between the arithmetic series and geometric series.
Complete step by step answer:
It is given, the geometric series is 3,-,--,--,768. And we have been asked to find the missing series.
We know that the geometric series is \[a,ar,a{{r}^{2}},a{{r}^{3}},,,,,a{{r}^{n-1}},a{{r}^{n}}\].
We have been given that, the fifth term is 768.
Comparing with the fourth term of geometric series, we get,
\[a{{r}^{4}}=768\] ………………………(1)
From the question, the first term is 3. That means a=3.
Now substitute ‘a’ in equation (1).
\[3{{r}^{4}}=768\] ……………………….(2)
Divide equation (2) by 3. We get,
\[\dfrac{3{{r}^{4}}}{3}=\dfrac{768}{3}\]
\[\Rightarrow {{r}^{4}}=256\]
We know that \[{{4}^{4}}=256\].
We get, \[{{r}^{4}}={{4}^{4}}\].
Taking cube root on both the sides,
\[\sqrt[4]{{{r}^{4}}}=\sqrt[4]{{{4}^{4}}}\]
Since \[\sqrt[4]{{{x}^{4}}}=x\], we get,
\[\therefore r=4\]
Therefore, we obtain the three missing terms as
The first term is
\[ar=3\left( 4 \right)\]
\[\therefore ar=12\].
The second term is
\[a{{r}^{2}}=3{{\left( 4 \right)}^{2}}\]
\[\Rightarrow a{{r}^{2}}=3\left( 16 \right)\]
\[\therefore a{{r}^{2}}=48\].
The third term is
\[a{{r}^{3}}=3{{\left( 4 \right)}^{3}}\]
\[\Rightarrow a{{r}^{3}}=3\left( 64 \right)\]
\[\therefore a{{r}^{3}}=192\]
The three missing terms are 12,48,192.
Hence, the complete geometric series is 3,12,48,192,768.
Note: Whenever we get this type of problem, we have to compare the given series with a geometric series. Arithmetic series should not be used for this type of question. Also, we should avoid calculation mistakes to obtain accurate answers. Avoid mistakes based on sign convention if any. Don’t get confused between the arithmetic series and geometric series.
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