The sum of three numbers in A.P. is 12, and the sum of their cubes is 408; find them.
Answer
Verified
Hint: Assume the numbers as (a – d), a, (a + d) and apply the conditions to solve to get the values of ‘a’ and ‘d’ and after that put that values in assumed numbers to get the answer.
Complete step-by-step answer:
As we have given that the three numbers are in A.P. and therefore we will assume (a-d), a, (a+d) as the three numbers in A.P. with ‘a’ as the first term and‘d’ is the common difference. As the sum of three numbers is 12, therefore we can write, (a - d) + a + (a + d) = 12 $\Rightarrow $a - d + a + a + d = 12 $\Rightarrow $a + a + a = 12 $\Rightarrow $ 3a = 12 $\Rightarrow a=\dfrac{12}{3}$ $\Rightarrow $ a = 4 …………………………………. (2) Now, as per the second condition given in the problem we can write, $\therefore {{\left( a-d \right)}^{3}}+{{a}^{3}}+{{\left( a+d \right)}^{3}}=408$ If we put the value equation (2) in the middle term of the above equation we will get, $\Rightarrow {{\left( a-d \right)}^{3}}+{{4}^{3}}+{{\left( a+d \right)}^{3}}=408$ $\Rightarrow {{\left( a-d \right)}^{3}}+64+{{\left( a+d \right)}^{3}}=408$ $\Rightarrow {{\left( a-d \right)}^{3}}+{{\left( a+d \right)}^{3}}=408-64$ $\Rightarrow {{\left( a-d \right)}^{3}}+{{\left( a+d \right)}^{3}}=344$ ………………………………… (3) Now to proceed further in the solution we should know the formula given below, Formula: \[\left( {{x}^{3}}+{{y}^{3}} \right)=\left( x+y \right)\times \left( {{x}^{2}}+{{y}^{2}}-xy \right)\] By using above formula we can write equation (3) as follows, \[\therefore \left[ \left( a-d \right)+\left( a+d \right) \right]\times \left[ {{\left( a-d \right)}^{2}}+{{\left( a+d \right)}^{2}}-\left( a-d \right)\times \left( a+d \right) \right]=344\] \[\therefore \left[ a-d+a+d \right]\times \left[ {{\left( a-d \right)}^{2}}+{{\left( a+d \right)}^{2}}-\left( a-d \right)\times \left( a+d \right) \right]=344\] \[\therefore \left[ a+a \right]\times \left[ {{\left( a-d \right)}^{2}}+{{\left( a+d \right)}^{2}}-\left( {{a}^{2}}+ad-ad-{{d}^{2}} \right) \right]=344\] \[\therefore 2a\times \left[ {{\left( a-d \right)}^{2}}+{{\left( a+d \right)}^{2}}-\left( {{a}^{2}}-{{d}^{2}} \right) \right]=344\] \[\therefore 2a\times \left[ {{\left( a-d \right)}^{2}}+{{\left( a+d \right)}^{2}}-{{a}^{2}}+{{d}^{2}} \right]=344\] Now to proceed further we should know the formulae given below, Formulae: \[{{\left( x+y \right)}^{2}}=\left( {{x}^{2}}+2xy+{{y}^{2}} \right)\] And \[{{\left( x+y \right)}^{2}}=\left( {{x}^{2}}-2xy+{{y}^{2}} \right)\] By using the above formulae we can write the above equation as, \[\therefore 2a\times \left[ \left( {{a}^{2}}-2ad+{{d}^{2}} \right)+\left( {{a}^{2}}+2ad+{{d}^{2}} \right)-{{a}^{2}}+{{d}^{2}} \right]=344\] By opening the brackets we will get, \[\Rightarrow 2a\times \left[ {{a}^{2}}-2ad+{{d}^{2}}+{{a}^{2}}+2ad+{{d}^{2}}-{{a}^{2}}+{{d}^{2}} \right]=344\] \[\Rightarrow 2a\times \left[ {{d}^{2}}+{{a}^{2}}+{{d}^{2}}+{{d}^{2}} \right]=344\] \[\Rightarrow 2a\times \left[ 3{{d}^{2}}+{{a}^{2}} \right]=344\] By substituting the value of equation (2) in the above equation we will get, \[\Rightarrow 2\times 4\times \left[ 3{{d}^{2}}+{{4}^{2}} \right]=344\] \[\Rightarrow 8\times \left[ 3{{d}^{2}}+{{4}^{2}} \right]=344\] \[{{\left( x+y \right)}^{3}}and{{\left( x-y \right)}^{3}}\] \[\Rightarrow 3{{d}^{2}}+16=43\] \[\Rightarrow 3{{d}^{2}}=43-16\] \[\Rightarrow 3{{d}^{2}}=27\] \[\Rightarrow {{d}^{2}}=\dfrac{27}{3}\] \[\Rightarrow {{d}^{2}}=9\] By taking square roots on both sides of the equation we will get, \[\therefore d=\pm 3\] Therefore d = 3 OR d = -3 ……………………………………. (4) Now we will rewrite the three numbers below, (a – d), a, (a + d) If we put the values of equation (2) and equation (4) in above equation as shown below, a = 4 and d = 3 Therefore numbers will become, (4 – 3), 4, (4+3) Therefore the numbers are, 1, 4, 7. Now, a = 4 and d = - 3 Therefore numbers will become, [4 – (-3)], 4, [4+(-3)] (4 + 3), 4, (4 – 3) Therefore the numbers are, 7, 4, 1. Therefore the three numbers are 1, 4, 7 or 7, 4, 1.
Note: Assume the standard numbers given by (a – d), a, (a + d) to make the calculations easier. Also in the step ${{\left( a-d \right)}^{3}}+{{\left( a+d \right)}^{3}}=344$ you can also use the formulae of \[{{\left( x+y \right)}^{3}}and{{\left( x-y \right)}^{3}}\]
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