# The sum of three numbers in AP is 12 and the sum of their cubes is 288. Find the numbers.

(a) 2, 4, 6 or 6, 4, 2

(b) 3, 4, 6 or 8, 4, 3

(c) 7, 8, 9 or 9, 8, 7

(d) 0, 4, 6 or 7, 4, 2

Last updated date: 26th Mar 2023

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Answer

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Hint: Take the second term of AP as a and common difference as d. Write three terms of AP and form equations relating terms of AP based on the data given in the question. Solve those equations to find the value of variables a and d. Substitute the value of variables to get the terms of AP.

Complete step-by-step answer:

We have three numbers in AP such that the sum of the numbers is 12 and the sum of cubes of the numbers is 288. We have to calculate the three numbers.

Let’s assume that the second term of the AP is a and the common difference of the terms is d.

Thus, we can write the other two terms as \[a-d\] and \[a+d\].

We know that the sum of these three numbers is 12. Thus, we have \[a-d+a+a+d=12\].

\[\begin{align}

& \Rightarrow 3a=12 \\

& \Rightarrow a=4 \\

\end{align}\]

We have to now find the value of common difference d.

We know that the sum of cubes of the numbers is 288. Thus, we have \[{{\left( a-d

\right)}^{3}}+{{a}^{3}}+{{\left( a+d \right)}^{3}}=288\].

We know that \[{{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}\] and

\[{{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}\].

Thus, we have

\[{{a}^{3}}-{{d}^{3}}-3{{a}^{2}}d+3a{{d}^{2}}+{{a}^{3}}+{{a}^{3}}+{{d}^{3}}+3{{a}^{2}}d+3a{{d}^{

2}}=288\].

Simplifying the above expression, we have \[3{{a}^{3}}+6a{{d}^{2}}=288\].

Further simplifying the equation, we have \[{{a}^{3}}+2a{{d}^{2}}=96\].

Substituting the value \[a=4\] in the above equation, we have \[{{\left( 4 \right)}^{3}}+2\left(

4 \right){{d}^{2}}=96\].

\[\begin{align}

& \Rightarrow 64+8{{d}^{2}}=96 \\

& \Rightarrow 8{{d}^{2}}=32 \\

& \Rightarrow {{d}^{2}}=4 \\

& \Rightarrow d=\pm 2 \\

\end{align}\]

Substituting the value \[a=4,d=\pm 2\], we have the terms of our AP as 6,4,2 or 2,4,6.

Hence, the terms of AP are 2,4,6 or 6,4,2, which is option (a).

Note: Arithmetic Progression is a sequence of numbers such that the difference between any two consecutive terms is a constant. One need not worry about getting two values of common difference and first term as they simply represent an increasing AP and a decreasing AP. We should also be careful while expanding the cubic power of equations.

Complete step-by-step answer:

We have three numbers in AP such that the sum of the numbers is 12 and the sum of cubes of the numbers is 288. We have to calculate the three numbers.

Let’s assume that the second term of the AP is a and the common difference of the terms is d.

Thus, we can write the other two terms as \[a-d\] and \[a+d\].

We know that the sum of these three numbers is 12. Thus, we have \[a-d+a+a+d=12\].

\[\begin{align}

& \Rightarrow 3a=12 \\

& \Rightarrow a=4 \\

\end{align}\]

We have to now find the value of common difference d.

We know that the sum of cubes of the numbers is 288. Thus, we have \[{{\left( a-d

\right)}^{3}}+{{a}^{3}}+{{\left( a+d \right)}^{3}}=288\].

We know that \[{{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}\] and

\[{{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}\].

Thus, we have

\[{{a}^{3}}-{{d}^{3}}-3{{a}^{2}}d+3a{{d}^{2}}+{{a}^{3}}+{{a}^{3}}+{{d}^{3}}+3{{a}^{2}}d+3a{{d}^{

2}}=288\].

Simplifying the above expression, we have \[3{{a}^{3}}+6a{{d}^{2}}=288\].

Further simplifying the equation, we have \[{{a}^{3}}+2a{{d}^{2}}=96\].

Substituting the value \[a=4\] in the above equation, we have \[{{\left( 4 \right)}^{3}}+2\left(

4 \right){{d}^{2}}=96\].

\[\begin{align}

& \Rightarrow 64+8{{d}^{2}}=96 \\

& \Rightarrow 8{{d}^{2}}=32 \\

& \Rightarrow {{d}^{2}}=4 \\

& \Rightarrow d=\pm 2 \\

\end{align}\]

Substituting the value \[a=4,d=\pm 2\], we have the terms of our AP as 6,4,2 or 2,4,6.

Hence, the terms of AP are 2,4,6 or 6,4,2, which is option (a).

Note: Arithmetic Progression is a sequence of numbers such that the difference between any two consecutive terms is a constant. One need not worry about getting two values of common difference and first term as they simply represent an increasing AP and a decreasing AP. We should also be careful while expanding the cubic power of equations.

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