Question & Answer

QUESTION

ANSWER
Verified

Hint: Assume that the first term of GP is ‘a’ and the common ratio is ‘r’. Write all the three terms of GP in terms of ‘a’ and ‘r’. Write the equations based on the data given in the question. Simplify the equations to calculate the value of ‘a’ and ‘r’. Substitute these values to calculate all the terms.

Complete step-by-step answer:

We have to find three terms of GP such that the sum of those terms is 21 and the sum of the square of terms is 189.

Let’s assume that the first term of GP is ‘a’ and the common ratio is ‘r’.

Thus, the three terms of GP are $a,ar,a{{r}^{2}}$.

We know that the sum of these three terms is 21. Thus, we have $a+ar+a{{r}^{2}}=21.....\left( 1 \right)$.

We know that the sum of the square of these terms is 189. Thus, we have ${{a}^{2}}+{{\left( ar \right)}^{2}}+{{\left( a{{r}^{2}} \right)}^{2}}=189.....\left( 2 \right)$.

Simplifying equation (1), we have $a\left( 1+r+{{r}^{2}} \right)=21$. Thus, we can rewrite this equation as $a=\dfrac{21}{1+r+{{r}^{2}}}.....\left( 3 \right)$.

Simplifying equation (2), we have ${{a}^{2}}+{{a}^{2}}{{r}^{2}}+{{a}^{2}}{{r}^{4}}=189$. Thus, we can rewrite this equation as ${{a}^{2}}\left( 1+{{r}^{2}}+{{r}^{4}} \right)=189\Rightarrow {{a}^{2}}=\dfrac{189}{1+{{r}^{2}}+{{r}^{4}}}.....\left( 4 \right)$.

Substituting equation (3) in equation (4), we have ${{\left( \dfrac{21}{1+r+{{r}^{2}}} \right)}^{2}}=\dfrac{189}{1+{{r}^{2}}+{{r}^{4}}}$.

Simplifying the above equation by dividing it by 63 on both sides, we have $\dfrac{7}{{{\left( 1+r+{{r}^{2}} \right)}^{2}}}=\dfrac{3}{1+{{r}^{2}}+{{r}^{4}}}.....\left( 5 \right)$.

We know the algebraic identity ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$.

Substituting $a=1,b=r,c={{r}^{2}}$ in the above equation, we have ${{\left( 1+r+{{r}^{2}} \right)}^{2}}=1+{{r}^{2}}+{{\left( {{r}^{2}} \right)}^{2}}+2r+2{{r}^{3}}+2{{r}^{2}}$.

Thus, we have ${{\left( 1+r+{{r}^{2}} \right)}^{2}}=1+3{{r}^{2}}+{{r}^{4}}+2r+2{{r}^{3}}.....\left( 6 \right)$.

Substituting equation (6) in equation (5), we have $\dfrac{7}{1+3{{r}^{2}}+{{r}^{4}}+2r+2{{r}^{3}}}=\dfrac{3}{1+{{r}^{2}}+{{r}^{4}}}$.

Cross multiplying the terms of the above equation, we have $7+7{{r}^{2}}+7{{r}^{4}}=3+6r+9{{r}^{2}}+6{{r}^{3}}+3{{r}^{4}}$.

Simplifying the above equation, we have $4{{r}^{4}}-6{{r}^{3}}-2{{r}^{2}}-6r+4=0$.

Dividing the above equation by 2, we have $2{{r}^{4}}-3{{r}^{3}}-{{r}^{2}}-3r+2=0$. We will now factorize this equation.

We can rewrite the above equation as $2{{r}^{4}}-4{{r}^{3}}+{{r}^{3}}-2{{r}^{2}}+{{r}^{2}}-2r-r+2=0$. Taking out the common terms, we have $2{{r}^{3}}\left( r-2 \right)+{{r}^{2}}\left( r-2 \right)+r\left( r-2 \right)-1\left( r-2 \right)=0$.

Thus, we have $\left( r-2 \right)\left( 2{{r}^{3}}+{{r}^{2}}+r-1 \right)=0$.

Further simplifying the above equation, we have $\left( r-2 \right)\left( 2{{r}^{3}}-{{r}^{2}}+2{{r}^{2}}-r+2r-1 \right)=0$.

Taking out the common terms, we have $\left( r-2 \right)\left\{ {{r}^{2}}\left( 2r-1 \right)+r\left( 2r-1 \right)+1\left( 2r-1 \right) \right\}=0$.

Thus, we have $\left( r-2 \right)\left( 2r-1 \right)\left( {{r}^{2}}+r+1 \right)=0$.

We will now factorize the equation ${{r}^{2}}+r+1=0$. Firstly, we will calculate the discriminant of this equation.

We know that the discriminant of the equation $a{{x}^{2}}+bx+c=0$ is ${{b}^{2}}-4ac$.

Substituting $a=1,b=1,c=1$ in the above expression, the discriminant of ${{r}^{2}}++r+1=0$ is $={{1}^{2}}-4\left( 1 \right)\left( 1 \right)=1-4=-3$.

As the discriminant of the equation ${{r}^{2}}++r+1=0$ is negative, this equation has imaginary roots.

Thus, the real roots of the equation $2{{r}^{4}}-3{{r}^{3}}-{{r}^{2}}-3r+2=0$ are the same as the roots of the equation $\left( r-2 \right)\left( 2r-1 \right)=0$.

So, we have $r-2=0$ or $2r-1=0$. Thus, we have $r=2,\dfrac{1}{2}.....\left( 7 \right)$.

We will now calculate the value of ‘a’.

Using equation (3), we have $a=\dfrac{21}{1+r+{{r}^{2}}}$. Substituting $r=2$ in the previous equation, we have $a=\dfrac{21}{1+2+{{2}^{2}}}=\dfrac{21}{1+2+4}=\dfrac{21}{7}=3$.

Thus, the terms of GP are $a,ar,a{{r}^{2}}=3,3\left( 2 \right),3{{\left( 2 \right)}^{2}}=3,6,12$.

Using equation (3), we have $a=\dfrac{21}{1+r+{{r}^{2}}}$. Substituting $r=\dfrac{1}{2}$ in the previous equation, we have $a=\dfrac{21}{1+\dfrac{1}{2}+{{\left( \dfrac{1}{2} \right)}^{2}}}=\dfrac{21}{1+\dfrac{1}{2}+\dfrac{1}{4}}=\dfrac{21}{\dfrac{4+2+1}{4}}=\dfrac{21\times 4}{7}=12$.

Thus, the terms of GP are $a,ar,a{{r}^{2}}=12,12\left( \dfrac{1}{2} \right),12{{\left( \dfrac{1}{2} \right)}^{2}}=12,6,3$.

Hence, the terms of GP which satisfy the given conditions are 3,6 and 12. or 12,6 and 3.

Note: Geometric Progression is a sequence of numbers in which the ratio of any two consecutive terms is a constant. We must observe that both the values of ‘a’ and ‘r’ give the same terms of the GP. One of them represents an increasing GP, while the other one represents a decreasing GP.

Complete step-by-step answer:

We have to find three terms of GP such that the sum of those terms is 21 and the sum of the square of terms is 189.

Let’s assume that the first term of GP is ‘a’ and the common ratio is ‘r’.

Thus, the three terms of GP are $a,ar,a{{r}^{2}}$.

We know that the sum of these three terms is 21. Thus, we have $a+ar+a{{r}^{2}}=21.....\left( 1 \right)$.

We know that the sum of the square of these terms is 189. Thus, we have ${{a}^{2}}+{{\left( ar \right)}^{2}}+{{\left( a{{r}^{2}} \right)}^{2}}=189.....\left( 2 \right)$.

Simplifying equation (1), we have $a\left( 1+r+{{r}^{2}} \right)=21$. Thus, we can rewrite this equation as $a=\dfrac{21}{1+r+{{r}^{2}}}.....\left( 3 \right)$.

Simplifying equation (2), we have ${{a}^{2}}+{{a}^{2}}{{r}^{2}}+{{a}^{2}}{{r}^{4}}=189$. Thus, we can rewrite this equation as ${{a}^{2}}\left( 1+{{r}^{2}}+{{r}^{4}} \right)=189\Rightarrow {{a}^{2}}=\dfrac{189}{1+{{r}^{2}}+{{r}^{4}}}.....\left( 4 \right)$.

Substituting equation (3) in equation (4), we have ${{\left( \dfrac{21}{1+r+{{r}^{2}}} \right)}^{2}}=\dfrac{189}{1+{{r}^{2}}+{{r}^{4}}}$.

Simplifying the above equation by dividing it by 63 on both sides, we have $\dfrac{7}{{{\left( 1+r+{{r}^{2}} \right)}^{2}}}=\dfrac{3}{1+{{r}^{2}}+{{r}^{4}}}.....\left( 5 \right)$.

We know the algebraic identity ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$.

Substituting $a=1,b=r,c={{r}^{2}}$ in the above equation, we have ${{\left( 1+r+{{r}^{2}} \right)}^{2}}=1+{{r}^{2}}+{{\left( {{r}^{2}} \right)}^{2}}+2r+2{{r}^{3}}+2{{r}^{2}}$.

Thus, we have ${{\left( 1+r+{{r}^{2}} \right)}^{2}}=1+3{{r}^{2}}+{{r}^{4}}+2r+2{{r}^{3}}.....\left( 6 \right)$.

Substituting equation (6) in equation (5), we have $\dfrac{7}{1+3{{r}^{2}}+{{r}^{4}}+2r+2{{r}^{3}}}=\dfrac{3}{1+{{r}^{2}}+{{r}^{4}}}$.

Cross multiplying the terms of the above equation, we have $7+7{{r}^{2}}+7{{r}^{4}}=3+6r+9{{r}^{2}}+6{{r}^{3}}+3{{r}^{4}}$.

Simplifying the above equation, we have $4{{r}^{4}}-6{{r}^{3}}-2{{r}^{2}}-6r+4=0$.

Dividing the above equation by 2, we have $2{{r}^{4}}-3{{r}^{3}}-{{r}^{2}}-3r+2=0$. We will now factorize this equation.

We can rewrite the above equation as $2{{r}^{4}}-4{{r}^{3}}+{{r}^{3}}-2{{r}^{2}}+{{r}^{2}}-2r-r+2=0$. Taking out the common terms, we have $2{{r}^{3}}\left( r-2 \right)+{{r}^{2}}\left( r-2 \right)+r\left( r-2 \right)-1\left( r-2 \right)=0$.

Thus, we have $\left( r-2 \right)\left( 2{{r}^{3}}+{{r}^{2}}+r-1 \right)=0$.

Further simplifying the above equation, we have $\left( r-2 \right)\left( 2{{r}^{3}}-{{r}^{2}}+2{{r}^{2}}-r+2r-1 \right)=0$.

Taking out the common terms, we have $\left( r-2 \right)\left\{ {{r}^{2}}\left( 2r-1 \right)+r\left( 2r-1 \right)+1\left( 2r-1 \right) \right\}=0$.

Thus, we have $\left( r-2 \right)\left( 2r-1 \right)\left( {{r}^{2}}+r+1 \right)=0$.

We will now factorize the equation ${{r}^{2}}+r+1=0$. Firstly, we will calculate the discriminant of this equation.

We know that the discriminant of the equation $a{{x}^{2}}+bx+c=0$ is ${{b}^{2}}-4ac$.

Substituting $a=1,b=1,c=1$ in the above expression, the discriminant of ${{r}^{2}}++r+1=0$ is $={{1}^{2}}-4\left( 1 \right)\left( 1 \right)=1-4=-3$.

As the discriminant of the equation ${{r}^{2}}++r+1=0$ is negative, this equation has imaginary roots.

Thus, the real roots of the equation $2{{r}^{4}}-3{{r}^{3}}-{{r}^{2}}-3r+2=0$ are the same as the roots of the equation $\left( r-2 \right)\left( 2r-1 \right)=0$.

So, we have $r-2=0$ or $2r-1=0$. Thus, we have $r=2,\dfrac{1}{2}.....\left( 7 \right)$.

We will now calculate the value of ‘a’.

Using equation (3), we have $a=\dfrac{21}{1+r+{{r}^{2}}}$. Substituting $r=2$ in the previous equation, we have $a=\dfrac{21}{1+2+{{2}^{2}}}=\dfrac{21}{1+2+4}=\dfrac{21}{7}=3$.

Thus, the terms of GP are $a,ar,a{{r}^{2}}=3,3\left( 2 \right),3{{\left( 2 \right)}^{2}}=3,6,12$.

Using equation (3), we have $a=\dfrac{21}{1+r+{{r}^{2}}}$. Substituting $r=\dfrac{1}{2}$ in the previous equation, we have $a=\dfrac{21}{1+\dfrac{1}{2}+{{\left( \dfrac{1}{2} \right)}^{2}}}=\dfrac{21}{1+\dfrac{1}{2}+\dfrac{1}{4}}=\dfrac{21}{\dfrac{4+2+1}{4}}=\dfrac{21\times 4}{7}=12$.

Thus, the terms of GP are $a,ar,a{{r}^{2}}=12,12\left( \dfrac{1}{2} \right),12{{\left( \dfrac{1}{2} \right)}^{2}}=12,6,3$.

Hence, the terms of GP which satisfy the given conditions are 3,6 and 12. or 12,6 and 3.

Note: Geometric Progression is a sequence of numbers in which the ratio of any two consecutive terms is a constant. We must observe that both the values of ‘a’ and ‘r’ give the same terms of the GP. One of them represents an increasing GP, while the other one represents a decreasing GP.