Question
Find four numbers in GP, whose sum is 85 and product is 4096.
✓
Answer
Let the four numbers in GP be
$\frac{a}{r^3}, \frac{a}{r}, a r, a r^3 \ldots$ (i)
Product of four numbers = 4096 [given]
$\begin{array}{l}\Rightarrow\left(\frac{a}{r^3}\right)\left(\frac{a}{r}\right)( ar )\left( ar ^3\right)=4096 \\ \Rightarrow a ^4=4096 \Rightarrow a ^4=8^4\end{array}$
On comparing the base of the power 4, we get
$\begin{array}{l}\Rightarrow \frac{a}{r^3}+\frac{a}{r}+ ar + ar ^3=85 \\ \Rightarrow a \left[\frac{1}{r^3}+\frac{1}{r}+r+r^3\right]=85 \\ \Rightarrow 8\left[ r ^3+\frac{1}{r^3}\right]+8\left[ r +\frac{1}{r}\right]=85[\because a =8] \\ \Rightarrow 8\left[\left(r+\frac{1}{r}\right)^3-3\left(r+\frac{1}{r}\right)\right]+8\left(r+\frac{1}{r}\right)=85\left[\because a ^3+ b ^3=( a + b )^3-3( a + b )\right] \\ \Rightarrow 8\left(r+\frac{1}{r}\right)^3-16\left(r+\frac{1}{r}\right)-85=0 \ldots \text { (ii) }\end{array}$
On putiing $\left(r+\frac{1}{r}\right)= x$ in Eq. (ii), we get
$8 x^3-16 x-85=0$
$\begin{array}{l}\Rightarrow(2 x-5)\left(4 x^2+10 x+17\right)=0 \\ \Rightarrow 2 x-5=0\left[\because 4 x^2+10 x+177=0 \text { has imaginary roots }\right] \\ \Rightarrow x=\frac{5}{2} \Rightarrow r+\frac{1}{r}=\frac{5}{2}\left[p u t x=r+\frac{1}{r}\right] \\ \Rightarrow 2 r^2-5 r+2=0 \\ \Rightarrow(r-2)(2 r-1)=0 \\ \Rightarrow r=2 \text { or } r=\frac{1}{2}\end{array}$
On putting $a =8$ and $r =2$ or $r =\frac{1}{2}$ in Eq. (i), we obtain four numbers as
$\frac{8}{2^3}, \frac{8}{2}, 8 \times 2,8 \times 2^3$
$\begin{array}{l}\text { or } \frac{8}{(1 / 2)^3}, \frac{8}{(1 / 2)}, 8 \times \frac{1}{2}, 8 \times\left(\frac{1}{2}\right)^3 \\ \Rightarrow 1,4,16,64 \text { or } 64,16,4,1\end{array}$
$\frac{a}{r^3}, \frac{a}{r}, a r, a r^3 \ldots$ (i)
Product of four numbers = 4096 [given]
$\begin{array}{l}\Rightarrow\left(\frac{a}{r^3}\right)\left(\frac{a}{r}\right)( ar )\left( ar ^3\right)=4096 \\ \Rightarrow a ^4=4096 \Rightarrow a ^4=8^4\end{array}$
On comparing the base of the power 4, we get
$\begin{array}{l}\Rightarrow \frac{a}{r^3}+\frac{a}{r}+ ar + ar ^3=85 \\ \Rightarrow a \left[\frac{1}{r^3}+\frac{1}{r}+r+r^3\right]=85 \\ \Rightarrow 8\left[ r ^3+\frac{1}{r^3}\right]+8\left[ r +\frac{1}{r}\right]=85[\because a =8] \\ \Rightarrow 8\left[\left(r+\frac{1}{r}\right)^3-3\left(r+\frac{1}{r}\right)\right]+8\left(r+\frac{1}{r}\right)=85\left[\because a ^3+ b ^3=( a + b )^3-3( a + b )\right] \\ \Rightarrow 8\left(r+\frac{1}{r}\right)^3-16\left(r+\frac{1}{r}\right)-85=0 \ldots \text { (ii) }\end{array}$
On putiing $\left(r+\frac{1}{r}\right)= x$ in Eq. (ii), we get
$8 x^3-16 x-85=0$
$\begin{array}{l}\Rightarrow(2 x-5)\left(4 x^2+10 x+17\right)=0 \\ \Rightarrow 2 x-5=0\left[\because 4 x^2+10 x+177=0 \text { has imaginary roots }\right] \\ \Rightarrow x=\frac{5}{2} \Rightarrow r+\frac{1}{r}=\frac{5}{2}\left[p u t x=r+\frac{1}{r}\right] \\ \Rightarrow 2 r^2-5 r+2=0 \\ \Rightarrow(r-2)(2 r-1)=0 \\ \Rightarrow r=2 \text { or } r=\frac{1}{2}\end{array}$
On putting $a =8$ and $r =2$ or $r =\frac{1}{2}$ in Eq. (i), we obtain four numbers as
$\frac{8}{2^3}, \frac{8}{2}, 8 \times 2,8 \times 2^3$
$\begin{array}{l}\text { or } \frac{8}{(1 / 2)^3}, \frac{8}{(1 / 2)}, 8 \times \frac{1}{2}, 8 \times\left(\frac{1}{2}\right)^3 \\ \Rightarrow 1,4,16,64 \text { or } 64,16,4,1\end{array}$
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