Question

The sum of the square of three consecutive natural numbers is 149. Find the numbers.

Answer 1

Hint: Consider three consecutive numbers as x, x + 1, x + 2. Find the sum of squares of these numbers and equate it to 149. Find the value of x and thus find the 3 consecutive numbers.

Complete step-by-step answer:

Let us consider the 3 consecutive numbers as x, x + 1, x + 2.

Given that the sum of squares of these 3 numbers is 149.

According to the question, we can say that,

\[{{x}^{2}}+{{\left( x+1 \right)}^{2}}+{{\left( x+2 \right)}^{2}}=149\]

We know, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\].

Expand the above expression and simplify it.

\[\begin{align}

  & {{x}^{2}}+{{x}^{2}}+2x+1+{{x}^{2}}+4x+4=149 \\

 & 3{{x}^{2}}+6x+5=149 \\

 & 3{{x}^{2}}+6x=149-5 \\

 & 3{{x}^{2}}+6x=144 \\

\end{align}\]

Divide the expression by 3.

\[\begin{align}

  & \Rightarrow {{x}^{2}}+2x=48 \\

 & \therefore {{x}^{2}}+2x-48=0 \\

\end{align}\]

The above expression is similar to the general equation, \[a{{x}^{2}}+bx+c=0\]. Now comparing both the equations we get,

a = 1, b = 2, c = -48.

Substitute these values in the quadratic formula.

\[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-2\pm \sqrt{{{\left( 2 \right)}^{2}}-4\times

1\times \left( -48 \right)}}{2\times 1}\]

\[=\dfrac{-2\pm \sqrt{4+192}}{2}=\dfrac{-2\pm \sqrt{196}}{2}=\dfrac{-2\pm 14}{2}\]

Thus the roots are \[\left( \dfrac{-2+14}{2} \right)\] = 6 and \[\left( \dfrac{-2-14}{2} \right)\] = -8.

Since, -8 is not a natural number, we will take x = 6.
Thus, when x = 6.

x + 1 = 6 + 1 = 7

x + 2 = 6 + 2 = 8

Thus we got the three natural numbers as 6, 7, 8.

Note: We have been asked to find a natural number, x = -8 is not a natural number, so we neglect that value. The natural numbers start from 1 and they correspond to positive integers. As it is said, take the consecutive number, take x, (x + 1) and (x + 2).