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The sum of the first three-term of a G.P is $\dfrac{39}{10}$ and their product is 1.Find the value of the number.
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Hint: In order to solve this question regarding the geometric progression of a series, we consider the three terms of the GP as $\dfrac{a}{r}, a, ar.$ and follow the steps as given in the question to get to our final result.
Complete step by step solution: A geometric progression(sequence) (also inaccurately known as a geometric series) is a sequence of numbers such that the quotient of any two successive members of the sequence is a constant called the common ratio of the sequence.
The geometric progression can be written as:
$\text{a}\text{r}^{0}=\text{a},\;\text{a}\text{r}^{1}=\text{ar},\;\text{a}\text{r}^{2},\; \text{a}\text{r}^{3},....$
where r ≠0, r is the common ratio and a is a scale factor(also the first term).
Let the first three-term of Geometric Progression be:-
$\dfrac{a}{r}$, a, ar
Then, according to the question; the sum of the first three terms is $=\dfrac{39}{10}.$.........given.
So,$\dfrac{a}{r}+\text{a}+\text{ar}= \dfrac{39}{10}$ ……...(i)
And the product of it's first three terms is 1, so we get;
$\Rightarrow \dfrac{a}{r}\times a\times ar=1$
$\Rightarrow {{a}^{3}}=1={{1}^{3}}$
$\Rightarrow ;\ \text{putting}\ $ in equation (i), we get;
$\Rightarrow a\left[ \dfrac{1}{r}+1+r \right]=\dfrac{39}{10}$
$\Rightarrow 1\left[ \dfrac{1}{r}+1+r \right]=\dfrac{39}{10}$
$\Rightarrow \left( \dfrac{1+r+{{r}^{2}}}{r} \right)=\dfrac{39}{10}$ $\Rightarrow 10\left( 1+r+{{r}^{2}} \right)=39\left( r \right)$
$\Rightarrow 10+10r+10{{r}^{2}}=39r$
$\Rightarrow 10+10r-39r+10{{r}^{2}}=0$
$\Rightarrow 10{{r}^{2}}-29r+10=0$
Now, we’ll solve this quadratic equation
$\Rightarrow 10{{r}^{2}}-25r-4r+10=0$
$\Rightarrow 5r\left( 2r-5 \right)-2\left( 2r-5 \right)=0$
$\Rightarrow \left( 2r-5 \right)\left( 5r-2 \right)=0$
$\left. \begin{matrix}
\text{So},2r-5=0 \\
\Rightarrow 2r=5 \\
\Rightarrow r=\dfrac{5}{2} \\
\end{matrix} \right|\begin{matrix}
\text{So,5r-2=0} \\
\Rightarrow 5r=2 \\
\Rightarrow r=\dfrac{2}{5} \\
\end{matrix}$
$\therefore r=\dfrac{5}{2},\dfrac{2}{5}.$
Now, for $a=1,\ \ \ \text{and}\ \ \ \text{r}\ \text{=}\ {2}/{5\ .}\;$
three term $=\dfrac{5}{2},1,\dfrac{2}{5}$ and for $a=1,\ \ \text{and}\ \ \ \text{r}\ \text{=}\ {5}/{2;}\;$three term $={}^{2}/{}_{5},1,{}^{5}/{}_{2}$
Note: A geometric progression(sequence) (also inaccurately known as a geometric series) is a sequence of numbers, such that the quotient of any two successive members of the sequence is a constant called the common ratio of the sequence.
The first three terms of the G.P can be represented as:-$\dfrac{a}{r},a,ar.$
The first five terms of the G.P can be represented as :- $\dfrac{a}{{{r}^{2}}},\dfrac{a}{r},a,ar,a{{r}^{2}}.$
Complete step by step solution: A geometric progression(sequence) (also inaccurately known as a geometric series) is a sequence of numbers such that the quotient of any two successive members of the sequence is a constant called the common ratio of the sequence.
The geometric progression can be written as:
$\text{a}\text{r}^{0}=\text{a},\;\text{a}\text{r}^{1}=\text{ar},\;\text{a}\text{r}^{2},\; \text{a}\text{r}^{3},....$
where r ≠0, r is the common ratio and a is a scale factor(also the first term).
Let the first three-term of Geometric Progression be:-
$\dfrac{a}{r}$, a, ar
Then, according to the question; the sum of the first three terms is $=\dfrac{39}{10}.$.........given.
So,$\dfrac{a}{r}+\text{a}+\text{ar}= \dfrac{39}{10}$ ……...(i)
And the product of it's first three terms is 1, so we get;
$\Rightarrow \dfrac{a}{r}\times a\times ar=1$
$\Rightarrow {{a}^{3}}=1={{1}^{3}}$
$\Rightarrow ;\ \text{putting}\ $ in equation (i), we get;
$\Rightarrow a\left[ \dfrac{1}{r}+1+r \right]=\dfrac{39}{10}$
$\Rightarrow 1\left[ \dfrac{1}{r}+1+r \right]=\dfrac{39}{10}$
$\Rightarrow \left( \dfrac{1+r+{{r}^{2}}}{r} \right)=\dfrac{39}{10}$ $\Rightarrow 10\left( 1+r+{{r}^{2}} \right)=39\left( r \right)$
$\Rightarrow 10+10r+10{{r}^{2}}=39r$
$\Rightarrow 10+10r-39r+10{{r}^{2}}=0$
$\Rightarrow 10{{r}^{2}}-29r+10=0$
Now, we’ll solve this quadratic equation
$\Rightarrow 10{{r}^{2}}-25r-4r+10=0$
$\Rightarrow 5r\left( 2r-5 \right)-2\left( 2r-5 \right)=0$
$\Rightarrow \left( 2r-5 \right)\left( 5r-2 \right)=0$
$\left. \begin{matrix}
\text{So},2r-5=0 \\
\Rightarrow 2r=5 \\
\Rightarrow r=\dfrac{5}{2} \\
\end{matrix} \right|\begin{matrix}
\text{So,5r-2=0} \\
\Rightarrow 5r=2 \\
\Rightarrow r=\dfrac{2}{5} \\
\end{matrix}$
$\therefore r=\dfrac{5}{2},\dfrac{2}{5}.$
Now, for $a=1,\ \ \ \text{and}\ \ \ \text{r}\ \text{=}\ {2}/{5\ .}\;$
three term $=\dfrac{5}{2},1,\dfrac{2}{5}$ and for $a=1,\ \ \text{and}\ \ \ \text{r}\ \text{=}\ {5}/{2;}\;$three term $={}^{2}/{}_{5},1,{}^{5}/{}_{2}$
Note: A geometric progression(sequence) (also inaccurately known as a geometric series) is a sequence of numbers, such that the quotient of any two successive members of the sequence is a constant called the common ratio of the sequence.
The first three terms of the G.P can be represented as:-$\dfrac{a}{r},a,ar.$
The first five terms of the G.P can be represented as :- $\dfrac{a}{{{r}^{2}}},\dfrac{a}{r},a,ar,a{{r}^{2}}.$
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