Question

Four numbers are in G.P. whose sum is 85 and the product is 4096, the numbers are:
(a) $1,16,64,4$
(b) $1,4,16,64$
(c) $64,1,4,16$
(d) $64,4,16,1$

Answer 1

Hint: For solving this question first we will assume 4 numbers in terms of two variables and then form some equations and solve the equations to get the correct answer.

Complete step-by-step solution -
Given:
Four numbers are in G.P. such that their product is 4096 and their addition is 85.
Now, before we solve this problem we should know that in a G.P. The ratio of consecutive terms is the same throughout the series and it is called the common ratio.
Now, let the four numbers of the G.P. are $\dfrac{a}{{{r}^{3}}},\dfrac{a}{r},ar,a{{r}^{3}}$ where ${{r}^{2}}$ is the common ratio. It is given in the question that the product of the four numbers in G.P. is 4096. Then,
$\begin{align}
  & \dfrac{a}{{{r}^{3}}}\times \dfrac{a}{r}\times ar\times a{{r}^{3}}=4096 \\
 & \Rightarrow {{a}^{4}}=4096 \\
 & \Rightarrow a={{\left( 4096 \right)}^{\dfrac{1}{4}}} \\
 & \Rightarrow a=8 \\
\end{align}$
Now, the four numbers are $\dfrac{8}{{{r}^{3}}},\dfrac{8}{r},8r,8{{r}^{3}}$ . It is given to us that their sum is 85. Then,
$\dfrac{8}{{{r}^{3}}}+\dfrac{8}{r}+8r+8{{r}^{3}}=85$
Now, we will solve the above equation to get the value of $r$ . Then,
$\begin{align}
  & \dfrac{8}{{{r}^{3}}}+\dfrac{8}{r}+8r+8{{r}^{3}}=85 \\
 & \Rightarrow 8+8{{r}^{2}}+8{{r}^{4}}+8{{r}^{6}}=85{{r}^{3}} \\
\end{align}$
$\Rightarrow 8{{r}^{6}}+8{{r}^{4}}-85{{r}^{3}}+8{{r}^{2}}+8=0$
Now, we will factorise the above equation and write the above equation as $8{{r}^{6}}-16{{r}^{5}}+16{{r}^{5}}-32{{r}^{4}}+40{{r}^{4}}-80{{r}^{3}}-5{{r}^{3}}+10{{r}^{2}}-2{{r}^{2}}+8$ . Then,
$\begin{align}
  & 8{{r}^{6}}-16{{r}^{5}}+16{{r}^{5}}-32{{r}^{4}}+40{{r}^{4}}-80{{r}^{3}}-5{{r}^{3}}+10{{r}^{2}}-2{{r}^{2}}+8=0 \\
 & \Rightarrow 8{{r}^{5}}\left( r-2 \right)+16{{r}^{4}}\left( r-2 \right)+40{{r}^{3}}\left( r-2 \right)-5{{r}^{2}}\left( r-2 \right)-2\left( {{r}^{2}}-4 \right)=0 \\
\end{align}$
Now, we can write ${{r}^{2}}-4=\left( r-2 \right)\left( r+2 \right)$ and take $\left( r-2 \right)$ common from every term. Then,
$\begin{align}
  & 8{{r}^{5}}\left( r-2 \right)+16{{r}^{4}}\left( r-2 \right)+40{{r}^{3}}\left( r-2 \right)-5{{r}^{2}}\left( r-2 \right)-2\left( {{r}^{2}}-4 \right)=0 \\
 & \Rightarrow 8{{r}^{5}}\left( r-2 \right)+16{{r}^{4}}\left( r-2 \right)+40{{r}^{3}}\left( r-2 \right)-5{{r}^{2}}\left( r-2 \right)-2\left( r-2 \right)\left( r+2 \right)=0 \\
 & \Rightarrow \left( r-2 \right)\left[ 8{{r}^{5}}+16{{r}^{4}}+40{{r}^{3}}-5{{r}^{2}}-2\left( r+2 \right) \right]=0 \\
\end{align}$
Now, we will factorise the above equation and write the above equation as $\left( r-2 \right)\left[ 8{{r}^{5}}-4{{r}^{4}}+20{{r}^{4}}-10{{r}^{3}}+50{{r}^{3}}-25{{r}^{2}}+20{{r}^{2}}-10r+8r-4 \right]=0$ . Then,
$\begin{align}
  & \left( r-2 \right)\left[ 8{{r}^{5}}-4{{r}^{4}}+20{{r}^{4}}-10{{r}^{3}}+50{{r}^{3}}-25{{r}^{2}}+20{{r}^{2}}-10r+8r-4 \right]=0 \\
 & \Rightarrow \left( r-2 \right)\left[ 4{{r}^{4}}\left( 2r-1 \right)+10{{r}^{3}}\left( 2r-1 \right)+25{{r}^{2}}\left( 2r-1 \right)+10r\left( 2r-1 \right)+4\left( 2r-1 \right) \right]=0 \\
\end{align}$
Now, we can take $\left( 2r-1 \right)$ common. Then,
$\begin{align}
  & \left( r-2 \right)\left[ 4{{r}^{4}}\left( 2r-1 \right)+10{{r}^{3}}\left( 2r-1 \right)+25{{r}^{2}}\left( 2r-1 \right)+10r\left( 2r-1 \right)+4\left( 2r-1 \right) \right]=0 \\
 & \Rightarrow \left( r-2 \right)\left( 2r-1 \right)\left[ 4{{r}^{4}}+10{{r}^{3}}+25{{r}^{2}}+10r+4 \right]=0 \\
\end{align}$
Now, in the above equation, we will have only two positive roots. As $4{{r}^{4}}+10{{r}^{3}}+25{{r}^{2}}+10r+4$ will not have any positive root because when we put any positive value of $r$ in it then it will be always greater than zero as in $4{{r}^{4}}+10{{r}^{3}}+25{{r}^{2}}+10r+4$ coefficients of ${{r}^{4}},{{r}^{3}},{{r}^{2}}\text{ and }r$ are positive and the constant term is also positive. And by going through options we can say that $r$ must be positive. Then, $\left( r-2 \right)$ or $\left( 2r-1 \right)$ will be zero. Then, $r=2,\dfrac{1}{2}$ .
When, $r=2$ and $a=8$ . Then, required four numbers will be:
$\begin{align}
  & \dfrac{a}{{{r}^{3}}}=\dfrac{8}{8}=1 \\
 & \dfrac{a}{r}=\dfrac{8}{2}=4 \\
 & ar=8\times 2=16 \\
 & a{{r}^{3}}=8\times 8=64 \\
\end{align}$
Thus, required four numbers are 1, 4, 16 and 64.
Now, when $r=\dfrac{1}{2}$ and $a=8$ . Then, the required numbers will be:
$\begin{align}
  & \dfrac{a}{{{r}^{3}}}=\dfrac{8}{{{\left( 0.5 \right)}^{3}}}=64 \\
 & \dfrac{a}{r}=\dfrac{8}{0.5}=16 \\
 & ar=8\times 0.5=4 \\
 & a{{r}^{3}}=8\times {{\left( 0.5 \right)}^{3}}=1 \\
\end{align}$
Thus, required four numbers are 64, 16, 4 and 1.
Hence, from all the above results we can say that required four numbers are 1, 4, 16 and 64.
Thus, (b) is the correct option.

Note: Here, the student must properly take the assumption of four numbers to avoid more calculation and solve the question without any calculation mistake. Moreover, even when we get the correct answer then, we should be careful while selecting the option. The four numbers should be in G.P. and their common ratio is either 4 or 0.25. For example, we cannot select option (a) 1, 16, 64, 4. Because numbers are not in G.P in the written sequence.