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10.  The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

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Let three terms of GP be a,ar,ar^2.

Then, we have   a+ar+ar^2=56

                         a(1+r+r^2)=56...............................................1

a-1,ar-7,ar^2-21   from an  AP.

\therefore ar-7-(a-1)=ar^2-21-(ar-7)

      ar-7-a+1=ar^2-21-ar+7

       ar-6-a=ar^2-14-ar                  

\Rightarrow ar^2-2ar+a=8

\Rightarrow ar^2-ar-ar+a=8
\Rightarrow a(r^2-2r+1)=8

\Rightarrow a(r^2-1)^2=8....................................................................2

From equation 1 and 2, we get 

\Rightarrow 7(r^2-2r+1)=1+r+r^2

\Rightarrow 7r^2-14r+7-1-r-r^2=0

\Rightarrow 6r^2-15r+6=0

\Rightarrow 2r^2-5r+2=0

\Rightarrow 2r^2-4r-r+2=0

\Rightarrow 2r(r-2)-1(r-2)=0

\Rightarrow (r-2)(2r-1)=0

    \Rightarrow r=2,r=\frac{1}{2}

If r=2, GP = 8,16,32

If r=0.2, GP= 32,16,8.

Thus, the numbers required are 8,16,32.

Posted by

seema garhwal

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