Answer

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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.

Let’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 GP’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.

Let’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 GP’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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