Question

The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is $87\dfrac{1}{2}$ .Find the largest of those numbers.

Answer 1

Hint: To find the largest of the three numbers, we have to consider the three numbers to be $\dfrac{a}{r},a,ar$ . We will then consider the given conditions. Multiplying these three numbers and equating them to 125 will result in an equation which can be solved to obtain the value of a. We have to consider the next given condition which will result in an equation of the form $\dfrac{a}{r}\times a+a\times ar+ar\times \dfrac{a}{r}=87\dfrac{1}{2}$ . We have to find the value of r from this. We have to substitute the value of a and r in the assumed numbers and find the largest among them.

Complete step by step solution:
Let us consider the three numbers in a Geometric Progression as $\dfrac{a}{r},a,ar$ . We are given that the product of three numbers in G.P. is 125.
$\Rightarrow \dfrac{a}{r}\times a\times ar=125$
Let us simplify the LHS by cancelling the common terms.
$\begin{align}
  & \Rightarrow a\times a\times a=125 \\
 & \Rightarrow {{a}^{3}}=125 \\
\end{align}$
We have to take the cube root of the above equation.
$\Rightarrow a=5$
We are also given that the sum of their products taken in pairs is $87\dfrac{1}{2}$ .
$\Rightarrow \dfrac{a}{r}\times a+a\times ar+ar\times \dfrac{a}{r}=87\dfrac{1}{2}$
We can write $87\dfrac{1}{2}$ as $\dfrac{175}{2}$ . Therefore, the above equation becomes
$\Rightarrow \dfrac{a}{r}\times a+a\times ar+ar\times \dfrac{a}{r}=\dfrac{175}{2}$
Let us simplify the LHS.
$\Rightarrow \dfrac{{{a}^{2}}}{r}+{{a}^{2}}r+{{a}^{2}}=\dfrac{175}{2}$
Let us take ${{a}^{2}}$ outside from LHS as it is common in each term.
$\Rightarrow {{a}^{2}}\left( \dfrac{1}{r}+r+1 \right)=\dfrac{175}{2}$
Now, we have to substitute the value of a in the above equation.
$\begin{align}
  & \Rightarrow {{5}^{2}}\left( \dfrac{1}{r}+r+1 \right)=\dfrac{175}{2} \\
 & \Rightarrow 25\left( \dfrac{1}{r}+r+1 \right)=\dfrac{175}{2} \\
\end{align}$
Let us take 25 to RHS.
$\begin{align}
  & \Rightarrow \dfrac{1}{r}+r+1=\dfrac{175}{2\times 25} \\
 & \Rightarrow \dfrac{1}{r}+r+1=\dfrac{7}{2} \\
\end{align}$
Let us take LCM of LHS and simplify.
$\Rightarrow \dfrac{1+{{r}^{2}}+r}{r}=\dfrac{7}{2}$
We have to take r to the RHS.
$\Rightarrow {{r}^{2}}+r+1=\dfrac{7r}{2}$
Let us bring the term in RHS to the LHS so that we can form an equation.
$\begin{align}
  & \Rightarrow {{r}^{2}}+r-\dfrac{7r}{2}+1=0 \\
 & \Rightarrow {{r}^{2}}+\dfrac{2r-7r}{2}+1=0 \\
 & \Rightarrow {{r}^{2}}-\dfrac{5r}{2}+1=0 \\
\end{align}$
Let us take the LCM.
$\begin{align}
  & \Rightarrow \dfrac{2{{r}^{2}}-5r+2}{2}=0 \\
 & \Rightarrow 2{{r}^{2}}-5r+2=0 \\
\end{align}$
Let us solve the above equation by splitting the middle term. We can write $-5r$ as $-4r-r$ .
$\Rightarrow 2{{r}^{2}}-4r-r+2=0$
We have to take the common terms outside.
$\Rightarrow 2r\left( r-2 \right)-1\left( r-2 \right)=0$
We can see that $r-2$ is common in LHS. Let us take it outside.
$\Rightarrow \left( 2r-1 \right)\left( r-2 \right)=0$
This means either $2r-1=0$ or $r-2=0$ .
Let us consider $2r-1=0$ . We have to solve for r.
$\begin{align}
  & \Rightarrow 2r-1=0 \\
 & \Rightarrow 2r=1 \\
 & \Rightarrow r=\dfrac{1}{2} \\
\end{align}$
Let us consider $r-2=0$ and solve for r.
\[\begin{align}
  & \Rightarrow r-2=0 \\
 & \Rightarrow r=2 \\
\end{align}\]
Hence, the value of r is $\dfrac{1}{2},2$ .
Hence, we can write the three numbers when $r=\dfrac{1}{2}$ as
 \[\begin{align}
  & \Rightarrow \dfrac{5}{\dfrac{1}{2}},5,5\times \dfrac{1}{2} \\
 & \Rightarrow 10,5,\dfrac{5}{2} \\
\end{align}\]
Similarly, the three numbers when $r=2$ are
$\begin{align}
  & \Rightarrow \dfrac{5}{2},5,5\times 2 \\
 & \Rightarrow \dfrac{5}{2},5,10 \\
\end{align}$
In either case, we can see that the largest number is 10.

Note: We have considered the three numbers to be $\dfrac{a}{r},a,ar$ because the question indicated that the numbers for a GP. We know that the ${{n}^{th}}$ term of a GP is given by $a{{r}^{n-1}}$ , where $n=0,1,2,...$ Hence, we can write the GP when substituting $n=0$ as
$a{{r}^{0-1}}=a{{r}^{-1}}=\dfrac{a}{r}$
Similarly, when $n=1$ , we will get a and when $n=2$ , we will get $a{{r}^{2}}$ and so on.