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If $2,7,9,5$ are subtracted respectively from four numbers forming a $\mathrm{GP}$, the resulting numbers are in $\mathrm{AP}$, then the smallest of the four numbers is
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Verified Answer
The correct answer is:
$-24$
Let the numbers be $a, a r, a r^{2}, a r^{3}$.
$\begin{array}{ll}
\Rightarrow & a-2, a r-7, a r^{2}-9 \text { are in A.P. } \\
\Rightarrow & a-2+a r^{2}-9=2(a r-7) \\
\Rightarrow & a\left(r^{2}+1\right)-11=2 a r-14 \\
\Rightarrow & r^{2}+1=2 r-\frac{3}{a}...(i)
\end{array}$
Further, $a r-7+a r^{3}-5=2\left(a r^{2}-9\right)$
$\Rightarrow \quad a r\left(r^{2}+1\right)-12=2 a r^{2}-18$
$\Rightarrow \quad r^{2}+1=2 r-\frac{6}{a r}...(ii)$
Solving Eqs. (i) and (ii), we get $r=2, a=-3$
The number are $-3,-6-12,-24$
$\begin{array}{ll}
\Rightarrow & a-2, a r-7, a r^{2}-9 \text { are in A.P. } \\
\Rightarrow & a-2+a r^{2}-9=2(a r-7) \\
\Rightarrow & a\left(r^{2}+1\right)-11=2 a r-14 \\
\Rightarrow & r^{2}+1=2 r-\frac{3}{a}...(i)
\end{array}$
Further, $a r-7+a r^{3}-5=2\left(a r^{2}-9\right)$
$\Rightarrow \quad a r\left(r^{2}+1\right)-12=2 a r^{2}-18$
$\Rightarrow \quad r^{2}+1=2 r-\frac{6}{a r}...(ii)$
Solving Eqs. (i) and (ii), we get $r=2, a=-3$
The number are $-3,-6-12,-24$
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