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Hint: Take any three numbers in GP such that the first term is a and common ratio is r. Write equations in terms of variables a and r based on the given data. Solve those equations to get the exact values of variables and thus, find the numbers.

Complete step-by-step answer:

We have to find three numbers in GP whose sum is 56 and if we subtract 1,7,21 from these numbers, we will obtain an AP.

Letâ€™s assume that the first term of the GP is a and the common ratio is r. Thus, the terms of the GP are \[a,ar,a{{r}^{2}}\]. We know that the sum of these terms is 56.

So, we have \[a+ar+a{{r}^{2}}=56.....\left( 1 \right)\].

Subtracting 1,7,21 from the terms of GP \[a,ar,a{{r}^{2}}\] in this order, we have the numbers \[a-1,ar-7,a{{r}^{2}}-21\].

We have the numbers \[a-1,ar-7,a{{r}^{2}}-21\] in AP.

We know that if three numbers x, y, z are in AP, we have \[2y=x+z\].

Substituting \[x=a-1,y=ar-7,z=a{{r}^{2}}-21\] in the above equation, we have \[2\left( ar-7 \right)=a-1+a{{r}^{2}}-21\].

Simplifying the above expression, we have \[2ar-14=a+a{{r}^{2}}-22\].

\[\Rightarrow a+a{{r}^{2}}=2ar+8.....\left( 2 \right)\]

Substituting equation (2) in equation (1), we have \[2ar+8+ar=56\].

Simplifying the above expression, we have \[3ar=48\Rightarrow ar=16\].

As \[ar=16\], we can write a as \[a=\dfrac{16}{r}\].

Substituting the equation \[a=\dfrac{16}{r}\] in equation (1), we have \[\dfrac{16}{r}\left( 1+r+{{r}^{2}} \right)=56\].

Simplifying the above expression, we have \[2+2r+2{{r}^{2}}=7r\].

\[\begin{align}

& \Rightarrow 2{{r}^{2}}-5r+2=0 \\

& \Rightarrow 2{{r}^{2}}-4r-r+2=0 \\

& \Rightarrow 2r\left( r-2 \right)-1\left( r-2 \right)=0 \\

& \Rightarrow \left( r-2 \right)\left( 2r-1 \right)=0 \\

& \Rightarrow r=2,\dfrac{1}{2} \\

\end{align}\]

Substituting the value \[r=2,\dfrac{1}{2}\] in the equation \[a=\dfrac{16}{r}\], we have \[a=8,32\].

Substituting \[r=2,a=8\] in the terms of GP \[a,ar,a{{r}^{2}}\], we have the terms of GP as 8,16,32.

Substituting \[r=\dfrac{1}{2},a=32\] in the terms of GP \[a,ar,a{{r}^{2}}\], we have the terms of GP as 32,16,8.

We observe that both the GPâ€™s are similar. One is an increasing GP, while the other one is a decreasing GP.

Hence, the terms of GP are 32, 16 and 8.

Note: Arithmetic Progression is a sequence of numbers such that the difference between any two consecutive terms is a constant. While, geometric progression is a sequence of numbers such that the ratio of any two consecutive terms is a constant. One need not worry about getting two values of common ratio and first term as they simply represent an increasing GP or a decreasing GP.

Complete step-by-step answer:

We have to find three numbers in GP whose sum is 56 and if we subtract 1,7,21 from these numbers, we will obtain an AP.

Letâ€™s assume that the first term of the GP is a and the common ratio is r. Thus, the terms of the GP are \[a,ar,a{{r}^{2}}\]. We know that the sum of these terms is 56.

So, we have \[a+ar+a{{r}^{2}}=56.....\left( 1 \right)\].

Subtracting 1,7,21 from the terms of GP \[a,ar,a{{r}^{2}}\] in this order, we have the numbers \[a-1,ar-7,a{{r}^{2}}-21\].

We have the numbers \[a-1,ar-7,a{{r}^{2}}-21\] in AP.

We know that if three numbers x, y, z are in AP, we have \[2y=x+z\].

Substituting \[x=a-1,y=ar-7,z=a{{r}^{2}}-21\] in the above equation, we have \[2\left( ar-7 \right)=a-1+a{{r}^{2}}-21\].

Simplifying the above expression, we have \[2ar-14=a+a{{r}^{2}}-22\].

\[\Rightarrow a+a{{r}^{2}}=2ar+8.....\left( 2 \right)\]

Substituting equation (2) in equation (1), we have \[2ar+8+ar=56\].

Simplifying the above expression, we have \[3ar=48\Rightarrow ar=16\].

As \[ar=16\], we can write a as \[a=\dfrac{16}{r}\].

Substituting the equation \[a=\dfrac{16}{r}\] in equation (1), we have \[\dfrac{16}{r}\left( 1+r+{{r}^{2}} \right)=56\].

Simplifying the above expression, we have \[2+2r+2{{r}^{2}}=7r\].

\[\begin{align}

& \Rightarrow 2{{r}^{2}}-5r+2=0 \\

& \Rightarrow 2{{r}^{2}}-4r-r+2=0 \\

& \Rightarrow 2r\left( r-2 \right)-1\left( r-2 \right)=0 \\

& \Rightarrow \left( r-2 \right)\left( 2r-1 \right)=0 \\

& \Rightarrow r=2,\dfrac{1}{2} \\

\end{align}\]

Substituting the value \[r=2,\dfrac{1}{2}\] in the equation \[a=\dfrac{16}{r}\], we have \[a=8,32\].

Substituting \[r=2,a=8\] in the terms of GP \[a,ar,a{{r}^{2}}\], we have the terms of GP as 8,16,32.

Substituting \[r=\dfrac{1}{2},a=32\] in the terms of GP \[a,ar,a{{r}^{2}}\], we have the terms of GP as 32,16,8.

We observe that both the GPâ€™s are similar. One is an increasing GP, while the other one is a decreasing GP.

Hence, the terms of GP are 32, 16 and 8.

Note: Arithmetic Progression is a sequence of numbers such that the difference between any two consecutive terms is a constant. While, geometric progression is a sequence of numbers such that the ratio of any two consecutive terms is a constant. One need not worry about getting two values of common ratio and first term as they simply represent an increasing GP or a decreasing GP.

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