# The continued product of three numbers in G.P is 216, and the sum of the product of them in pairs is 156. Find the numbers.

Last updated date: 18th Mar 2023

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Answer

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Hint: First of all take three numbers in G.P as \[\dfrac{a}{r},a,ar\]. Then use the given information that the product of these numbers is equal to 216 to find the value of a. Further, use the value of the sum of the product of them in pairs to get the values of r. Hence, find the numbers.

Complete step by step solution:

We are given that the continued product of three numbers in G.P is 216 and the sum of the product of that in pairs is 156. We have to find the numbers.

First of all, we know that the general term of G.P is \[a{{r}^{n-1}}\]. So, let us consider three terms in G.P as,

\[\dfrac{a}{r},a,ar\]

We are given that the product of these three numbers in G.P is 216. So, we get

\[\dfrac{a}{r}.a.ar=216\]

By canceling the like terms, we get,

\[{{a}^{3}}=216\]

By taking the cube root on both sides of the above equation, we get,

\[\Rightarrow a=\sqrt[3]{216}\]

Or, a = 6

Now by substituting the value of a = 6 in terms \[\dfrac{a}{r},a,ar\]. We get three terms as,

\[\dfrac{6}{r},6,6r\]

Also, we are given that the sum of the products of each pair of the given terms is 156. So, we get,

\[\left( \dfrac{6}{r} \right).\left( 6 \right)+\left( 6 \right).\left( 6r \right)+\left( 6r \right).\left( \dfrac{6}{r} \right)=156\]

By simplifying the above equation and taking 6.6 = 36 common, we get,

\[36\left( \dfrac{1}{r}+r+\dfrac{r}{r} \right)=156\]

Or, \[36\left( \dfrac{1+{{r}^{2}}+r}{r} \right)=156\]

By dividing both sides by 12, we get,

\[3\left( \dfrac{1+{{r}^{2}}+r}{r} \right)=13\]

By cross multiplying the above equation, we get,

\[3+3{{r}^{2}}+3r=13r\]

Or, \[3{{r}^{2}}+3r-13r+3=0\]

\[\Rightarrow 3{{r}^{2}}-10r+3=0\]

We can write 10r = 9r + r. So, we get,

\[3{{r}^{2}}-9r-r+3=0\]

We can also write the above equation as,

\[3r\left( r-3 \right)-1\left( r-3 \right)=0\]

By taking (r – 3) common, we get,

\[\left( r-3 \right)\left( 3r-1 \right)=0\]

Therefore, we get r = 3 and \[r=\dfrac{1}{3}\].

By substituting the value of r = 3 in terms \[\dfrac{6}{r},6,6r\]. We get three terms as

\[\dfrac{6}{3},6,6\times 3\]

\[=2,6,18\]

Also, by substituting the value of \[r=\dfrac{1}{3}\] in terms \[\dfrac{6}{r},6,6r\]. We get three terms as,

\[=\dfrac{6}{\dfrac{1}{3}},6,6.\left( \dfrac{1}{3} \right)\]

\[=18,6,2\]

Hence, we get three numbers in G.P as 2, 6, 18, or 18, 6, 2.

Note: Students should always take 3 numbers in G.P as \[\dfrac{a}{r},a,ar\] to easily solve the given problem. Also, students often make this mistake of writing three terms as a, ar and \[a{{r}^{2}}\] after getting the values of a and r. But they must note that as we have taken the terms as \[\dfrac{a}{r},a,ar\]. So they must substitute a and r in these terms to get 3 numbers in G.P. So, this mistake must be avoided.

Complete step by step solution:

We are given that the continued product of three numbers in G.P is 216 and the sum of the product of that in pairs is 156. We have to find the numbers.

First of all, we know that the general term of G.P is \[a{{r}^{n-1}}\]. So, let us consider three terms in G.P as,

\[\dfrac{a}{r},a,ar\]

We are given that the product of these three numbers in G.P is 216. So, we get

\[\dfrac{a}{r}.a.ar=216\]

By canceling the like terms, we get,

\[{{a}^{3}}=216\]

By taking the cube root on both sides of the above equation, we get,

\[\Rightarrow a=\sqrt[3]{216}\]

Or, a = 6

Now by substituting the value of a = 6 in terms \[\dfrac{a}{r},a,ar\]. We get three terms as,

\[\dfrac{6}{r},6,6r\]

Also, we are given that the sum of the products of each pair of the given terms is 156. So, we get,

\[\left( \dfrac{6}{r} \right).\left( 6 \right)+\left( 6 \right).\left( 6r \right)+\left( 6r \right).\left( \dfrac{6}{r} \right)=156\]

By simplifying the above equation and taking 6.6 = 36 common, we get,

\[36\left( \dfrac{1}{r}+r+\dfrac{r}{r} \right)=156\]

Or, \[36\left( \dfrac{1+{{r}^{2}}+r}{r} \right)=156\]

By dividing both sides by 12, we get,

\[3\left( \dfrac{1+{{r}^{2}}+r}{r} \right)=13\]

By cross multiplying the above equation, we get,

\[3+3{{r}^{2}}+3r=13r\]

Or, \[3{{r}^{2}}+3r-13r+3=0\]

\[\Rightarrow 3{{r}^{2}}-10r+3=0\]

We can write 10r = 9r + r. So, we get,

\[3{{r}^{2}}-9r-r+3=0\]

We can also write the above equation as,

\[3r\left( r-3 \right)-1\left( r-3 \right)=0\]

By taking (r – 3) common, we get,

\[\left( r-3 \right)\left( 3r-1 \right)=0\]

Therefore, we get r = 3 and \[r=\dfrac{1}{3}\].

By substituting the value of r = 3 in terms \[\dfrac{6}{r},6,6r\]. We get three terms as

\[\dfrac{6}{3},6,6\times 3\]

\[=2,6,18\]

Also, by substituting the value of \[r=\dfrac{1}{3}\] in terms \[\dfrac{6}{r},6,6r\]. We get three terms as,

\[=\dfrac{6}{\dfrac{1}{3}},6,6.\left( \dfrac{1}{3} \right)\]

\[=18,6,2\]

Hence, we get three numbers in G.P as 2, 6, 18, or 18, 6, 2.

Note: Students should always take 3 numbers in G.P as \[\dfrac{a}{r},a,ar\] to easily solve the given problem. Also, students often make this mistake of writing three terms as a, ar and \[a{{r}^{2}}\] after getting the values of a and r. But they must note that as we have taken the terms as \[\dfrac{a}{r},a,ar\]. So they must substitute a and r in these terms to get 3 numbers in G.P. So, this mistake must be avoided.

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