Answer
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Hint: Here, we will take the three numbers as \[a - d\], \[a\] and \[a + d\]. Then we will use the given conditions of the question to the value of \[a\] and \[d\]. Then we will substitute the values of \[a\] and \[d\] in the assumed three numbers to find the numbers.
Complete step-by-step answer:
It is given that the sum of three numbers in AP is 21 and the product of these three numbers is 231.
We know that these three numbers are in AP.
Let us assume that these three required numbers are \[a - d\], \[a\] and \[a + d\].
Adding the above three numbers, we get
\[
\Rightarrow a - d + a + a + d \\
\Rightarrow 3a \\
\]
Since we know that the above sum equals to 21.
Taking the above sum equals to 21, we get
\[3a = 21\]
Dividing the above equation by 3 on each if the sides, we get
\[
\Rightarrow \dfrac{{3a}}{3} = \dfrac{{21}}{3} \\
\Rightarrow a = 7 \\
\]
Finding the product of these three numbers, we get
\[\left( {a - d} \right)a\left( {a + d} \right) = a\left( {a - d} \right)\left( {a + d} \right)\]
Using the property \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] in the above equation, we get
\[a\left( {{a^2} - {d^2}} \right)\]
Taking the above product is equals to 231, we get
\[a\left( {{a^2} - {d^2}} \right) = 231\]
Substituting the value of \[a\] in the above equation, we get
\[
\Rightarrow 7\left( {{7^2} - {d^2}} \right) = 231 \\
\Rightarrow 7\left( {49 - {d^2}} \right) = 231 \\
\Rightarrow 343 - 7{d^2} = 231 \\
\Rightarrow 7{d^2} = 343 - 231 \\
\Rightarrow 7{d^2} = 112 \\
\]
Dividing the above equation by 7 on each side, we get
\[
\Rightarrow \dfrac{{7{d^2}}}{7} = \dfrac{{112}}{7} \\
\Rightarrow {d^2} = 16 \\
\]
Taking the square root in the above equation on each side, we get
\[ \Rightarrow d = \pm 4\]
\[ \Rightarrow d = 4\] or \[d = - 4\]
Replacing 4 for \[d\] and 7 for \[a\] in the three numbers assumed are \[a - d\], \[a\] and \[a + d\], we get
\[7 - 4 = 3\]
7
\[7 + 4 = 11\]
Replacing \[ - 4\] for \[d\] and 7 for \[a\] in the three numbers assumed are \[a - d\], \[a\] and \[a + d\], we get
\[
7 - \left( { - 4} \right) = 7 + 4 \\
= 11 \\
\]
7
\[
7 + \left( { - 4} \right) = 7 - 4 \\
= 3 \\
\]
Hence, the required numbers are 3, 7, 11 or 11, 7, 3.
Note: In solving these types of questions, we have to assume the middle number is equal to any variable and then we will assume the first and second number such that they are less and more than the assumed middle number by a common difference, \[d\]. Then when we add the three numbers we will get the value of the middle number and after that, we can easily find the other two numbers.
Complete step-by-step answer:
It is given that the sum of three numbers in AP is 21 and the product of these three numbers is 231.
We know that these three numbers are in AP.
Let us assume that these three required numbers are \[a - d\], \[a\] and \[a + d\].
Adding the above three numbers, we get
\[
\Rightarrow a - d + a + a + d \\
\Rightarrow 3a \\
\]
Since we know that the above sum equals to 21.
Taking the above sum equals to 21, we get
\[3a = 21\]
Dividing the above equation by 3 on each if the sides, we get
\[
\Rightarrow \dfrac{{3a}}{3} = \dfrac{{21}}{3} \\
\Rightarrow a = 7 \\
\]
Finding the product of these three numbers, we get
\[\left( {a - d} \right)a\left( {a + d} \right) = a\left( {a - d} \right)\left( {a + d} \right)\]
Using the property \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] in the above equation, we get
\[a\left( {{a^2} - {d^2}} \right)\]
Taking the above product is equals to 231, we get
\[a\left( {{a^2} - {d^2}} \right) = 231\]
Substituting the value of \[a\] in the above equation, we get
\[
\Rightarrow 7\left( {{7^2} - {d^2}} \right) = 231 \\
\Rightarrow 7\left( {49 - {d^2}} \right) = 231 \\
\Rightarrow 343 - 7{d^2} = 231 \\
\Rightarrow 7{d^2} = 343 - 231 \\
\Rightarrow 7{d^2} = 112 \\
\]
Dividing the above equation by 7 on each side, we get
\[
\Rightarrow \dfrac{{7{d^2}}}{7} = \dfrac{{112}}{7} \\
\Rightarrow {d^2} = 16 \\
\]
Taking the square root in the above equation on each side, we get
\[ \Rightarrow d = \pm 4\]
\[ \Rightarrow d = 4\] or \[d = - 4\]
Replacing 4 for \[d\] and 7 for \[a\] in the three numbers assumed are \[a - d\], \[a\] and \[a + d\], we get
\[7 - 4 = 3\]
7
\[7 + 4 = 11\]
Replacing \[ - 4\] for \[d\] and 7 for \[a\] in the three numbers assumed are \[a - d\], \[a\] and \[a + d\], we get
\[
7 - \left( { - 4} \right) = 7 + 4 \\
= 11 \\
\]
7
\[
7 + \left( { - 4} \right) = 7 - 4 \\
= 3 \\
\]
Hence, the required numbers are 3, 7, 11 or 11, 7, 3.
Note: In solving these types of questions, we have to assume the middle number is equal to any variable and then we will assume the first and second number such that they are less and more than the assumed middle number by a common difference, \[d\]. Then when we add the three numbers we will get the value of the middle number and after that, we can easily find the other two numbers.
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