CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 65 Marks
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 $]$
$\Rightarrow\left(\frac{a}{r^3}\right)\left(\frac{a}{r}\right)( ar )\left( ar ^3\right)=4096$
$\Rightarrow a ^4=4096$
$\Rightarrow a ^4=8^4$
On comparing the base of the power $4,$ we get
$\Rightarrow \frac{a}{r^3}+\frac{a}{r}+ ar + ar ^3=85$
$\Rightarrow a \left[\frac{1}{r^3}+\frac{1}{r}+r+r^3\right]=85$
$\Rightarrow 8\left[ r ^3+\frac{1}{r^3}\right]+8\left[ r +\frac{1}{r}\right]=85[\because a =8]$
$\Rightarrow 8\left[\left(r+\frac{1}{r}\right)^3-3\left(r+\frac{1}{r}\right)\right]+8\left(r+\frac{1}{r}\right)=85\left[\because a ^3+ b ^3=( a + b )^3-3( a + b )\right]$
$\Rightarrow 8\left(r+\frac{1}{r}\right)^3-16\left(r+\frac{1}{r}\right)-85=0 \ldots \text { (ii) }$
On putiing $\left(r+\frac{1}{r}\right)= x$ in Eq. $(ii),$ we get
$8 x^3-16 x-85=0$
$\Rightarrow(2 x-5)\left(4 x^2+10 x+17\right)=0$
$\Rightarrow 2 x-5=0 [\because 4 x^2+10 x+177=0$ has imaginary roots $]$
$\Rightarrow x=\frac{5}{2}$
$\Rightarrow r+\frac{1}{r}=\frac{5}{2} [$ put $x=r+\frac{1}{r}]$
$\Rightarrow 2 r^2-5 r+2=0$
$\Rightarrow(r-2)(2 r-1)=0$
$\Rightarrow r=2$ or $r=\frac{1}{2}$
On putting $a =8$ and $r =2$ or $r =\frac{1}{2}$ in Eq. $(i),$
we obtain four numbers as $\frac{8}{2^3}, \frac{8}{2}, 8 \times 2,8 \times 2^3$
$\text { or } \frac{8}{(1 / 2)^3}, \frac{8}{(1 / 2)}, 8 \times \frac{1}{2}, 8 \times\left(\frac{1}{2}\right)^3$
$\Rightarrow 1,4,16,64$ or $64,16,4,1$
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