Find three numbers in GP, whose product is 8 and their sum is \[\dfrac{{26}}{3}\]
Answer
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Hint: Here in this question, we have to find the sum of finite geometric series. The geometric series is defined as the series with a constant ratio between the two successive terms. Then by considering the geometric series we have found the sum of the series.
Complete step by step solution:
In mathematics we have three types of series namely, arithmetic series, geometric series and harmonic series.
The geometric series is defined as the series with a constant ratio between the two successive terms. The finite geometric series is generally represented as \[a,ar,a{r^2},...,a{r^n}\], where a is first term and r is a common ratio.
The first three terms of GP is generally given as \[\dfrac{a}{r}\], a and \[ar\]
The product of three terms is 8. So we have
\[\dfrac{a}{r} \times a \times ar = 8\]
On multiplying we have
\[ \Rightarrow {a^3} = 8\]
On applying the cubic root to the above equation we have
\[ \Rightarrow a = 2\]
As we know that the sum of three terms is \[\dfrac{{26}}{3}\]
So we have
\[\dfrac{a}{r} + a + ar = \dfrac{{26}}{3}\]
On substituting the value of a, the above equation is written as
\[ \Rightarrow \dfrac{2}{r} + 2 + 2r = \dfrac{{26}}{3}\]
On simplifying we have
\[
\Rightarrow 2 + 2r + 2{r^2} = \dfrac{{26}}{3}r \\
\Rightarrow 6 + 6r + 6{r^2} = 26r \\
\]
Take 26r to LHS of the equation
\[
\Rightarrow 6{r^2} + 6r - 26r + 6 = 0 \\
\Rightarrow 6{r^2} - 20r + 6 = 0 \\
\]
Divide the above equation by 3
\[ \Rightarrow 3{r^2} - 10r + 3 = 0\]
The equation can be written as
\[
\Rightarrow 3{r^2} - 9r - r + 3 = 0 \\
\Rightarrow 3r(r - 3) - 1(r - 3) = 0 \\
\Rightarrow (r - 3)(3r - 1) = 0 \\
\]
\[ \Rightarrow r = 3\]and \[r = \dfrac{1}{3}\]
When a = 2 and r = 3, the geometric sequence is \[\dfrac{2}{3},2,6\]
When a = 2 and r = \[\dfrac{1}{3}\], the geometric sequence is \[6,2,\dfrac{2}{3}\]
Note: Three different forms of series are arithmetic series, geometric series and harmonic series. For the arithmetic series is the series with common differences. The geometric series is the series with a common ratio. The sum is known as the total value of the given series.
Complete step by step solution:
In mathematics we have three types of series namely, arithmetic series, geometric series and harmonic series.
The geometric series is defined as the series with a constant ratio between the two successive terms. The finite geometric series is generally represented as \[a,ar,a{r^2},...,a{r^n}\], where a is first term and r is a common ratio.
The first three terms of GP is generally given as \[\dfrac{a}{r}\], a and \[ar\]
The product of three terms is 8. So we have
\[\dfrac{a}{r} \times a \times ar = 8\]
On multiplying we have
\[ \Rightarrow {a^3} = 8\]
On applying the cubic root to the above equation we have
\[ \Rightarrow a = 2\]
As we know that the sum of three terms is \[\dfrac{{26}}{3}\]
So we have
\[\dfrac{a}{r} + a + ar = \dfrac{{26}}{3}\]
On substituting the value of a, the above equation is written as
\[ \Rightarrow \dfrac{2}{r} + 2 + 2r = \dfrac{{26}}{3}\]
On simplifying we have
\[
\Rightarrow 2 + 2r + 2{r^2} = \dfrac{{26}}{3}r \\
\Rightarrow 6 + 6r + 6{r^2} = 26r \\
\]
Take 26r to LHS of the equation
\[
\Rightarrow 6{r^2} + 6r - 26r + 6 = 0 \\
\Rightarrow 6{r^2} - 20r + 6 = 0 \\
\]
Divide the above equation by 3
\[ \Rightarrow 3{r^2} - 10r + 3 = 0\]
The equation can be written as
\[
\Rightarrow 3{r^2} - 9r - r + 3 = 0 \\
\Rightarrow 3r(r - 3) - 1(r - 3) = 0 \\
\Rightarrow (r - 3)(3r - 1) = 0 \\
\]
\[ \Rightarrow r = 3\]and \[r = \dfrac{1}{3}\]
When a = 2 and r = 3, the geometric sequence is \[\dfrac{2}{3},2,6\]
When a = 2 and r = \[\dfrac{1}{3}\], the geometric sequence is \[6,2,\dfrac{2}{3}\]
Note: Three different forms of series are arithmetic series, geometric series and harmonic series. For the arithmetic series is the series with common differences. The geometric series is the series with a common ratio. The sum is known as the total value of the given series.
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