Three numbers are in GP whose sum is 70 . if the extremes be each multiplied by 4 and mean by 5 , they will be in AP . find the numbers.

Solution:    Let the three numbers in GP be  $\frac{a}{r},a,ar$

Given    ,                 $\frac{a}{r}+a+ar=70$

and    $\frac{4a}{r},5a,4ar$  are in AP.

$\therefore$                              $10a=\frac{4a}{r}+4ar\Rightarrow \frac{10a}{4}=\frac{a}{r}+ar$

or                                 $\frac{5a}{2}=70-a$          From  Eq.(1)

or                             $5a=140-2a\Rightarrow a=20.$

From Equation (1)  , we get

$\frac{20}{r}+20+20r=70$   or    $\frac{20}{r}+20r=50$

$\Rightarrow$                           $2r^{2}-5r+2=0\Rightarrow (r-2)(2r-1)=0$

$\therefore$                                           $r=2,\frac{1}{2}$

Hence , the three numbers are $10,20,40$  or      $40,20,10.$

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