
Insert four GMs between \[3\] and \[96\].
Answer
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Hint: The geometric mean of any two positive numbers \[a\] and \[b\] is the number \[\sqrt {ab} \], such that the sequence \[a,G,b\] is G.P. Here, in the given question, we have to insert four geometric means (GMs) in between \[3\] and \[96\] which implies the resultant Geometric Progression (G.P.) will be of total six terms. Now, at first, we will assume the four unknown values to form a G.P. and then we will use the formula of the general term of a G.P. to calculate the common ratio \[r\]. As we have common ration now, we will use it to calculate the four required GMs.
Formula used:
Geometric Mean of any two positive numbers \[a\] and \[b\]= \[\sqrt {ab} \]
General term of a G.P. = \[a{r^{n - 1}}\]
Complete step-by-step solution:
Let \[{G_1},{G_2},{G_3},{G_4}\] be the four numbers between \[3\] and \[96\] such that \[3,{G_1},{G_2},{G_3},{G_4},96\] is a G.P.
Now, we know that,
General term of a G.P. = \[a{r^{n - 1}}\]
Then, the sixth term of a G.P. = \[a{r^5}\]
In the above G.P, \[96\] is the sixth term, and \[a = 3\] is the first term, putting that value,
\[96 = 3{r^5}\]
Simplifying it, we get,
\[ {r^5} = 32 \\
\therefore r = 2 \]
Now, using this \[r\] and general term formula, we will calculate the four required GMs,
General term of a G.P. = \[a{r^{n - 1}}\]
Second term of G.P. is calculated by \[{G_1} = ar\]
\[
{G_1} = 3\left( 2 \right) \\
\therefore {G_1} = 6 \\
\]
Third term of the G.P. is calculated by, \[{G_2} = a{r^2}\]
\[
{G_2} = 3{\left( 2 \right)^2} \\
\therefore {G_2} = 12 \\
\]
Fourth term of a G.P. is calculated by, \[{G_3} = a{r^3}\]
\[
{G_3} = 3{\left( 2 \right)^3} \\
\therefore {G_3} = 24 \\
\]
Fifth term of G.P. is calculated by, \[{G_4} = a{r^4}\]
\[
{G_4} = 3{(2)^4} \\
\therefore {G_4} = 48 \\
\]
Hence, we can insert \[6,12,24,48\] between \[3\] and \[96\] so that the resulting sequences are in G.P.
Note: Some of the interesting properties of the Geometric Mean are: (1). Geometric mean of any given data set will always be less than or equal to the arithmetic mean of the same data set. (2). If each and every term of a G.P. is substituted by the G.M., then the product of the terms remains unchanged. (3). The ratio and the product of the corresponding terms of the G.M. in two series are equal to the ratio and the product of their geometric mean respectively.
Formula used:
Geometric Mean of any two positive numbers \[a\] and \[b\]= \[\sqrt {ab} \]
General term of a G.P. = \[a{r^{n - 1}}\]
Complete step-by-step solution:
Let \[{G_1},{G_2},{G_3},{G_4}\] be the four numbers between \[3\] and \[96\] such that \[3,{G_1},{G_2},{G_3},{G_4},96\] is a G.P.
Now, we know that,
General term of a G.P. = \[a{r^{n - 1}}\]
Then, the sixth term of a G.P. = \[a{r^5}\]
In the above G.P, \[96\] is the sixth term, and \[a = 3\] is the first term, putting that value,
\[96 = 3{r^5}\]
Simplifying it, we get,
\[ {r^5} = 32 \\
\therefore r = 2 \]
Now, using this \[r\] and general term formula, we will calculate the four required GMs,
General term of a G.P. = \[a{r^{n - 1}}\]
Second term of G.P. is calculated by \[{G_1} = ar\]
\[
{G_1} = 3\left( 2 \right) \\
\therefore {G_1} = 6 \\
\]
Third term of the G.P. is calculated by, \[{G_2} = a{r^2}\]
\[
{G_2} = 3{\left( 2 \right)^2} \\
\therefore {G_2} = 12 \\
\]
Fourth term of a G.P. is calculated by, \[{G_3} = a{r^3}\]
\[
{G_3} = 3{\left( 2 \right)^3} \\
\therefore {G_3} = 24 \\
\]
Fifth term of G.P. is calculated by, \[{G_4} = a{r^4}\]
\[
{G_4} = 3{(2)^4} \\
\therefore {G_4} = 48 \\
\]
Hence, we can insert \[6,12,24,48\] between \[3\] and \[96\] so that the resulting sequences are in G.P.
Note: Some of the interesting properties of the Geometric Mean are: (1). Geometric mean of any given data set will always be less than or equal to the arithmetic mean of the same data set. (2). If each and every term of a G.P. is substituted by the G.M., then the product of the terms remains unchanged. (3). The ratio and the product of the corresponding terms of the G.M. in two series are equal to the ratio and the product of their geometric mean respectively.
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