Question

Three consecutive positive integers are such that the square of their sum exceeds the sum of the square by 214. Which are those three integers?
A). 6, 7, 8
B). 4, 5, 6
C). 7, 8, 9
D). 5, 6, 7

Answer 1

Hint: Let us assume the first positive integer as n, and then the next two consecutive positive integers will be n+1 and n+2. Now, it is given that the square of the sum of these three consecutive integers exceeds the sum of squares of each of the three consecutive integers by 214 so write the sum of three consecutive integers and then square it. After that, we are going to equate this square of the summation to the sum of the square of the three consecutive integers and 214. Now, solve this equation and find the value of n and hence, will find the three consecutive integers.

Complete step-by-step solution
Let us consider three positive consecutive integers as n, n+1 and n+2. Now, taking the sum of these three consecutive positive integers we get,
$\begin{align}
  & n+n+1+n+2 \\
 & =3n+3 \\
\end{align}$
Squaring the above summation will give us:
${{\left( 3n+3 \right)}^{2}}$ ………… (1)
Now, squaring the three consecutive integers and after squaring each of them we are going to add them.
${{\left( n \right)}^{2}}+{{\left( n+1 \right)}^{2}}+{{\left( n+2 \right)}^{2}}$ ………. (2)
It is given above that the square of the sum of three consecutive positive integers exceeds the sum of the square of each of the three positive integers by 214 so we are adding 214 in (2).
${{\left( n \right)}^{2}}+{{\left( n+1 \right)}^{2}}+{{\left( n+2 \right)}^{2}}+214$………….. (3)
Equating (1) and (3) we get,
${{\left( 3n+3 \right)}^{2}}={{\left( n \right)}^{2}}+{{\left( n+1 \right)}^{2}}+{{\left( n+2 \right)}^{2}}+214$……… (4)
We know that,
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Using this relation in solving (4) we get,
$\begin{align}
  & {{\left( 3n+3 \right)}^{2}}={{\left( n \right)}^{2}}+{{\left( n+1 \right)}^{2}}+{{\left( n+2 \right)}^{2}}+214 \\
 & \Rightarrow 9{{n}^{2}}+18n+9={{n}^{2}}+{{n}^{2}}+2n+1+{{n}^{2}}+4n+4+214 \\
 & \Rightarrow 9{{n}^{2}}+18n+9=3{{n}^{2}}+6n+219 \\
 & \Rightarrow 6{{n}^{2}}+12n-210=0 \\
\end{align}$
Dividing 6 on both the sides we get,
$\begin{align}
  & \dfrac{6{{n}^{2}}+12n-210}{6}=0 \\
 & \Rightarrow {{n}^{2}}+2n-35=0 \\
\end{align}$
Factoring the above quadratic equation by writing 2n as 7n – 5n we get,
${{n}^{2}}+7n-5n-35=0$
Taking n as common from the first two terms and -5 as common from the last two terms of the above expression we get,
$\begin{align}
  & n\left( n+7 \right)-5\left( n+7 \right)=0 \\
 & \Rightarrow \left( n+7 \right)\left( n-5 \right)=0 \\
\end{align}$
Equating each bracket to 0 we get,
$\begin{align}
  & n+7=0 \\
 & n=-7 \\
 & n-5=0 \\
 & \Rightarrow n=5 \\
\end{align}$
In the above we have assumed n as a positive integer so -7 will get rejected and the value of n is equal to 5.
Now, we got the value of n so we can find the value of three consecutive positive integers which is equal to:
$\begin{align}
  & n,n+1,n+2 \\
 & =5,5+1,5+2 \\
 & =5,6,7 \\
\end{align}$
Hence, the correct option is (d).

Note: The alternative approach to solve the above problem is by checking the options given above and then seeing which of the 4 options are satisfying the condition that the square of the sum of these three consecutive integers exceeds the sum of squares of each of the three consecutive integers by 214.
Let us check option (a)6, 7, 8
Adding the three integers followed by squaring their summation.
$\begin{align}
  & {{\left( 6+7+8 \right)}^{2}} \\
 & =441..........(5) \\
\end{align}$
Now, squaring each of the three integers (6, 7, 8) and then adding them we get,
$\begin{align}
  & {{\left( 6 \right)}^{2}}+{{\left( 7 \right)}^{2}}+{{\left( 8 \right)}^{2}} \\
 & =36+49+64 \\
 & =149.........(6) \\
\end{align}$
It is given that the square of the sum of three consecutive positive integers exceeds the sum of the square of each of the three positive integers by 214 so we are adding 214 in (6).
$\begin{align}
  & 149+214 \\
 & =363.........(7) \\
\end{align}$
Now, according to the condition above, (5) and (7) should be equal but as you can see they are not so option (a) is incorrect.
Similarly, you can check all the options and find the correct answer.