Question

The sum of the four terms in G.P is 820 and their product is 5,31,441. Find the numbers?

Answer 1

Hint: We start solving the problem by assigning the variables for the first term and common ratio for the given G.P (Geometric Progression). We then use our first condition that the sum of the four terms in G.P is 820 to get our first equation. We then use our second condition that the product of four terms is 531441 to get the value of the first terms in terms of common ratio. We these two relations and makes subsequent arrangements and calculations to get the value of the first term and common ratio. Using these, we find the required four numbers.

Complete step-by-step solution:
According to the problem, we have four numbers in G.P (Geometric Progression) whose sum is 820, and the product is 5,31,441. We need to find all four numbers.
Let us assume the first term of required G.P (Geometric Progression) is ‘a’ and the common ration be ‘r’.
We know that the next term in a G.P (Geometric Progression) can be found by multiplying the preceding term with a common ratio. Using this we get our four terms as $a$, $ar$, $a{{r}^{2}}$ and $a{{r}^{3}}$.
We have sum of the four terms in G.P as 820. So, we get $a+ar+a{{r}^{2}}+a{{r}^{3}}=820$ ---(1).
We have product of the four terms in G.P as 5,31,441. So, we get \[a\times \left( ar \right)\times \left( a{{r}^{2}} \right)\times \left( a{{r}^{3}} \right)=531441\].
\[\Rightarrow {{a}^{4}}{{r}^{6}}=531441\].
\[\Rightarrow {{a}^{4}}=\dfrac{531441}{{{r}^{6}}}\].
\[\Rightarrow a=\dfrac{\sqrt[4]{531441}}{{{\left( {{r}^{6}} \right)}^{\dfrac{1}{4}}}}\].
\[\Rightarrow a=\dfrac{27}{{{r}^{\dfrac{3}{2}}}}\] ---(2).
Let us substitute equation (2) in equation (1).
$\Rightarrow a+ar+a{{r}^{2}}+a{{r}^{3}}=820$.
$\Rightarrow a\left( 1+r+{{r}^{2}}+{{r}^{3}} \right)=820$.
$\Rightarrow \dfrac{27}{{{r}^{\dfrac{3}{2}}}}\times \left( 1+r+{{r}^{2}}+{{r}^{3}} \right)=820$.
$\Rightarrow \left( \dfrac{1}{{{r}^{\dfrac{3}{2}}}}+\dfrac{r}{{{r}^{\dfrac{3}{2}}}}+\dfrac{{{r}^{2}}}{{{r}^{\dfrac{3}{2}}}}+\dfrac{{{r}^{3}}}{{{r}^{\dfrac{3}{2}}}} \right)=\dfrac{820}{27}$.
$\Rightarrow \left( \dfrac{1}{{{r}^{\dfrac{3}{2}}}}+\dfrac{1}{{{r}^{\dfrac{1}{2}}}}+{{r}^{\dfrac{1}{2}}}+{{r}^{\dfrac{3}{2}}} \right)=\dfrac{820}{27}$.
$\Rightarrow \left( {{\left( \dfrac{1}{{{r}^{\dfrac{1}{2}}}} \right)}^{3}}+{{\left( {{r}^{\dfrac{1}{2}}} \right)}^{3}} \right)+\left( \dfrac{1}{{{r}^{\dfrac{1}{2}}}}+{{r}^{\dfrac{1}{2}}} \right)=\dfrac{820}{27}$.
Let us assume ${{r}^{\dfrac{1}{2}}}=x$ ---(3).
\[\Rightarrow \left( {{\left( \dfrac{1}{x} \right)}^{3}}+{{\left( x \right)}^{3}} \right)+\left( \dfrac{1}{x}+x \right)=\dfrac{820}{27}\].
We can see that \[{{\left( \dfrac{1}{x} \right)}^{3}}+{{\left( x \right)}^{3}}\] resembles ${{a}^{3}}+{{b}^{3}}$ which is equal to $\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$.
\[\Rightarrow \left( \left( \dfrac{1}{x}+x \right)\left( {{\left( \dfrac{1}{x} \right)}^{2}}+{{\left( x \right)}^{2}}-\left( \left( \dfrac{1}{x} \right)\left( x \right) \right) \right) \right)+\left( \dfrac{1}{x}+x \right)=\dfrac{820}{27}\].
\[\Rightarrow \left( \left( \dfrac{1}{x}+x \right)\left( {{\left( \dfrac{1}{x} \right)}^{2}}+{{\left( x \right)}^{2}}-1 \right) \right)+\left( \dfrac{1}{x}+x \right)=\dfrac{820}{27}\].
\[\Rightarrow \left( \dfrac{1}{x}+x \right)\left( {{\left( \dfrac{1}{x} \right)}^{2}}+{{\left( x \right)}^{2}}-1+1 \right)=\dfrac{820}{27}\].
\[\Rightarrow \left( \dfrac{1}{x}+x \right)\left( {{\left( \dfrac{1}{x} \right)}^{2}}+{{\left( x \right)}^{2}} \right)=\dfrac{820}{27}\].
\[\Rightarrow \left( \dfrac{1}{x}+x \right)\left( {{\left( \dfrac{1}{x}+x \right)}^{2}}-2\left( \dfrac{1}{x} \right)\left( x \right) \right)=\dfrac{820}{27}\].
\[\Rightarrow \left( \dfrac{1}{x}+x \right)\left( {{\left( \dfrac{1}{x}+x \right)}^{2}}-2 \right)=\dfrac{820}{27}\].
\[\Rightarrow {{\left( \dfrac{1}{x}+x \right)}^{3}}-2\left( \dfrac{1}{x}+x \right)=\dfrac{820}{27}\].
Let us assume \[\left( \dfrac{1}{x}+x \right)=y\] ---(4).
\[\Rightarrow {{y}^{3}}-2y=\dfrac{820}{27}\].
\[\Rightarrow 27\left( {{y}^{3}}-2y \right)=820\].
\[\Rightarrow 27{{y}^{3}}-54y-820=0\].
\[\Rightarrow 27{{y}^{3}}-90{{y}^{2}}+90{{y}^{2}}-300y+246y-820=0\].
\[\Rightarrow 9{{y}^{2}}\left( 3y-10 \right)+30y\left( 3y-10 \right)+82\left( 3y-10 \right)=0\].
\[\Rightarrow \left( 3y-10 \right)\left( 9{{y}^{2}}+30y+82 \right)=0\].
$\Rightarrow 3y-10=0$.
$\Rightarrow 3y=10$.
$\Rightarrow y=\dfrac{10}{3}$.
Let us find the discriminant of \[9{{y}^{2}}+30y+82=0\]. So, discriminant is $D={{30}^{2}}-4\left( 9 \right)\left( 82 \right)$.
$\Rightarrow D=900-2952$.
$\Rightarrow D=-2052$.
We got the discriminant less than zero which means \[9{{y}^{2}}+30y+82=0\] doesn’t have real roots. SO, we have only real root for \[27{{y}^{3}}-54y-820=0\] as $y=\dfrac{10}{3}$ ---(5).
Let us substitute equation (5) in equation (4).
$\Rightarrow \dfrac{1}{x}+x=\dfrac{10}{3}$.
$\Rightarrow \dfrac{1+{{x}^{2}}}{x}=\dfrac{10}{3}$.
$\Rightarrow 3\left( 1+{{x}^{2}} \right)=10\left( x \right)$.
$\Rightarrow 3+3{{x}^{2}}=10x$.
$\Rightarrow 3{{x}^{2}}-10x+3=0$.
$\Rightarrow 3{{x}^{2}}-9x-x+3=0$.
$\Rightarrow 3x\left( x-3 \right)-1\left( x-3 \right)=0$.
$\Rightarrow \left( 3x-1 \right)\left( x-3 \right)=0$.
$\Rightarrow \left( 3x-1 \right)=0 or \left( x-3 \right)=0$.
$\Rightarrow 3x=1 or x=3$.
$\Rightarrow x=\dfrac{1}{3} or x=3$.
Let us take the value of x as 3. We substitute this equation (3).
$\Rightarrow {{r}^{\dfrac{1}{2}}}=x$.
$\Rightarrow {{r}^{\dfrac{1}{2}}}=3$.
$\Rightarrow r={{3}^{2}}$.
$\Rightarrow r=9$---(6).
Let us substitute equation (6) in equation (2).
$\Rightarrow a=\dfrac{27}{{{r}^{\dfrac{3}{2}}}}$.
$\Rightarrow a=\dfrac{27}{{{9}^{\dfrac{3}{2}}}}$.
$\Rightarrow a=\dfrac{27}{27}$.
$\Rightarrow a=1$.
We have got the values of first term and common ratio as 1 and 9.
Let us find the terms of G.P
First term = $a=1$.
Second term = $ar=\left( 1\times 9 \right)=9$.
Third term = $a{{r}^{2}}=\left( 1\times {{9}^{2}} \right)=81.$
Fourth term = $a{{r}^{3}}=\left( 1\times {{9}^{3}} \right)=729$.
We have got the four terms of G.P as 1, 9, 81, 729.
∴ The required four terms of G.P are 1, 9, 81, 729.
We can also take the value of x as $\dfrac{1}{3}$ to solve for the values of the first term and common ratio respectively as follows:
$\Rightarrow {{r}^{\dfrac{1}{2}}}=x$.
$\Rightarrow {{r}^{\dfrac{1}{2}}}=\dfrac{1}{3}$.
$\Rightarrow r={{\left( \dfrac{1}{3} \right)}^{2}}$.
$\Rightarrow r=\dfrac{1}{9}$.
Let us find the value of ‘a’.
$\Rightarrow a=\dfrac{27}{{{r}^{\dfrac{3}{2}}}}$.
$\Rightarrow a=\dfrac{27}{{{\left( \dfrac{1}{9} \right)}^{\dfrac{3}{2}}}}$.
$\Rightarrow a=\dfrac{27}{\dfrac{1}{27}}$.
$\Rightarrow a=27\times 27$.
$\Rightarrow a=729$.
We have got the values of the first term and common ratio as 729 and $\dfrac{1}{9}$.
Let us find the terms of G.P
First term = $a=729$.
Second term = $ar=\left( 729\times \dfrac{1}{9} \right)=81$.
Third term = $a{{r}^{2}}=\left( 729\times {{\left( \dfrac{1}{9} \right)}^{2}} \right)=9.$
Fourth term = $a{{r}^{3}}=\left( 1\times {{\left( \dfrac{1}{9} \right)}^{3}} \right)=1$.
We just found the same numbers in decreasing order.

Note: We should not confuse the product as 5, 31 and 441 separately. We should not forget that the discriminant of the quadratic equation should be greater than or equal to zero in order to have roots. We can also find extend the geometric progression series using the terms and common ratio we just obtained. We can see in the case of $r=\dfrac{1}{9}$, the terms of the progression are in decreasing order for which we can find the finite sum of infinite terms.