Question

Find three numbers in G.P. whose sum is $ \dfrac{13}{12} $ and products is -1.

Answer 1

Hint: To solve the question given above, we will first find out what is the full form of G.P and what kind of terms are present in G.P. Then, we will assume that the first term of the given G.P is a and the constant ratio is r. Thus, we will get the following three terms: $ a,ar,a{{r}^{2}} $ . Then, we will multiply the three terms and we will equate it to -1. From here, we will write a in terms of r. Another equation is obtained by adding the terms and equating it to $ \dfrac{13}{12} $ . In this equation, we will put the value of a in terms of r. From here, we will get values of a and r and hence we will get the required numbers of G.P.

Complete step-by-step answer:
Before we start to solve the question, we must know what is a G.P. The full G.P is a geometric progression. Geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non zero number called the common ratio.
Let us assume that the fixed term of the G.P is ‘a’ and the common ratio is ‘r’. Thus, we will get the following three numbers: $ a,ar,a{{r}^{2}} $ . Now, it is given in question that the product of three numbers is -1. Thus, we will get:
 $ \begin{align}
  & a\times ar\times a{{r}^{2}}=-1 \\
 & \Rightarrow {{a}^{3}}{{r}^{3}}={{\left( -1 \right)}^{3}} \\
\end{align} $
On taking cube root on both sides, we will get,
 $ \begin{align}
  & \Rightarrow \sqrt[3]{{{a}^{3}}{{r}^{3}}}=\left( -1 \right) \\
 & \Rightarrow ar=-1 \\
 & \Rightarrow a=\dfrac{-1}{r}............\left( 1 \right) \\
\end{align} $
Now, it is given in question that the sum of three numbers is $ \dfrac{13}{12} $ . Thus, we have:
 $ \begin{align}
  & a+ar+a{{r}^{2}}=\dfrac{13}{12} \\
 & a\left( 1+r+{{r}^{2}} \right)=\dfrac{13}{12} \\
\end{align} $
Now, we will put the value of ‘a’ from (1) to the above equation. Thus, we will get:
 $ \begin{align}
  & \left( \dfrac{-1}{r} \right)\left( 1+r+{{r}^{2}} \right)=\dfrac{13}{12} \\
 & \Rightarrow 12\left( 1+r+{{r}^{2}} \right)=-13r \\
 & \Rightarrow 12{{r}^{2}}+12r+12=-13r \\
 & \Rightarrow 12{{r}^{2}}+25r+12=0 \\
\end{align} $
Now, we will solve the above quadratic equation by factorisation. Thus, we have:
 $ \begin{align}
  & \Rightarrow 12{{r}^{2}}+16r+9r+12=0 \\
 & \Rightarrow 4r\left( 3r+4 \right)+3\left( 3r+4 \right)=0 \\
 & \Rightarrow \left( 4r+3 \right)\left( 3r+4 \right)=0 \\
 & \Rightarrow 4r+3=0\ or\ \Rightarrow 3r+4=0 \\
 & \Rightarrow 4r=-3\ or\ \Rightarrow 3r=-4 \\
 & \Rightarrow r=\dfrac{-3}{4}\ or\ \Rightarrow r=\dfrac{-4}{3} \\
\end{align} $
Now, we will take any value of r as both the values of r will give same three numbers. Let us take $ r=\dfrac{-3}{4} $ . Thus, we will put this value of r in (1).
 $ \begin{align}
  & \Rightarrow a=\dfrac{-1}{\left( \dfrac{-3}{4} \right)} \\
 & \Rightarrow a=\dfrac{4}{3} \\
\end{align} $
Thus, the required numbers in G.P will be,
  $ \begin{align}
  & \Rightarrow \dfrac{4}{3},\dfrac{4}{3}\left( \dfrac{-3}{4} \right),\dfrac{4}{3}{{\left( \dfrac{-3}{4} \right)}^{2}} \\
 & \Rightarrow \text{Required numbers }\Rightarrow \dfrac{4}{3},-1,\dfrac{3}{4} \\
\end{align} $

Note: We can also solve the question in an alternate way as shown:
Let the middle number be a and common ratio be r. Thus, the numbers will be : $ \dfrac{a}{r},a\ and\ ar $ . Now, their product is (-1). So,
 $ \begin{align}
  & \dfrac{a}{r}\times a\times ar=-1 \\
 & \Rightarrow {{a}^{3}}=-1 \\
 & \Rightarrow a=-1 \\
\end{align} $
Now, their sum is $ \dfrac{13}{12} $ . Thus,
 $ \begin{align}
  & \dfrac{a}{r}+a+ar=\dfrac{13}{12} \\
 & \Rightarrow \dfrac{-1}{r}-1-r=\dfrac{13}{12} \\
 & \Rightarrow 12\left( {{r}^{2}}+r+1 \right)=-13 \\
 & \Rightarrow 12{{r}^{2}}+25r+12=0 \\
 & \Rightarrow \left( 4r+3 \right)\left( 3r+4 \right)=0 \\
 & \Rightarrow r=\dfrac{-3}{4},\dfrac{4}{3} \\
\end{align} $
Required numbers $ =\dfrac{4}{3},-1,\dfrac{3}{4}\ or\ \dfrac{3}{4},-1,\dfrac{4}{3} $ .