Question

The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the sum of their squares.

Answer 1

Hint:For the above question, we will assume the numbers in A.P. as a, a+d, a+2d, where d is the common difference of the A.P. Now, we can formulate equations using the data: the sum of numbers is 12 and the sum of their cubes is 288. Solving these will give us the terms a and d.

Complete step-by-step answer:
A.P. is also called as arithmetic progression in which the difference of the consecutive terms are equal and known as the common difference of the A.P.
We have been given that the sum of numbers in A.P. is 12 and the sum of their cubes is 288.
Let us suppose the three numbers in A.P. are as a, a + d, a + 2d, where ‘a’ is the first term and ‘d’ is the common difference of the A.P.
$a+\left( a+d \right)+\left( a+2d \right)=12$
On simplifying the above equation, we get,

& 3\left( a+d \right)=12 \\\ & a+d=4 \\\ & a=4-d......(1) \\\ \end{aligned}$$ Again, $${{a}^{3}}+{{\left( a+d \right)}^{3}}+{{\left( a+2d \right)}^{3}}=288.....(2)$$ By substituting the value of ‘a’ from equation (1) to equation (2), we get as follows: $${{\left( 4-d \right)}^{3}}+{{\left( 4 \right)}^{3}}+{{\left( 4+d \right)}^{3}}=288$$ Hence, we will use the formula to expand the cubic equation as shown below: $$\begin{aligned} & {{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b) \\\ & {{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b) \\\ & \Rightarrow {{4}^{3}}-{{d}^{3}}-12d(4-d)+{{4}^{3}}+{{4}^{3}}+{{d}^{3}}+12d(4+d)=288 \\\ \end{aligned}$$ On further simplification, we get the above equation as follows: $$24{{d}^{2}}=288-192$$ On dividing by 24 to both sides of equality, we get as follows: $$\begin{aligned} & \dfrac{24}{24}{{d}^{2}}=\dfrac{96}{24} \\\ & \therefore {{d}^{2}}=4 \\\ \end{aligned}$$ Taking square root both sides of equality, we get as follows: $$\therefore {{d}^{2}}=\pm 2$$ So we get the common difference d of the A.P. is $$\pm 2$$. For d = 2, $$\begin{aligned} & a=4-d \\\ & a=4-2=2 \\\ & \therefore a=2 \\\ \end{aligned}$$ So the numbers will be 2, 4 and 6. For d = -2, $$\begin{aligned} & a=4-(-2) \\\ & \therefore a=6 \\\ \end{aligned}$$ The numbers will be 6, 4 and 2. But the sum of the squares of the term of A.P. will be the same as shown below. $${{2}^{2}}+{{4}^{2}}+{{6}^{2}}={{6}^{2}}+{{4}^{2}}+{{2}^{2}}=56$$ Therefore, the sum of the squares of the term is equal to 56. Note: Be careful while expanding the cubic term and use the cubic identity properly. Don’t miss the second value of common difference i.e. -2. Whichever value is taken, we will get the same numbers, just the order of those changes.For these types of problems we can also assume three numbers of an A.P are $a-d$ ,$a$, $a+d$ in which we get the same answer.