Question

The sum of three consecutive integers is 72. What is the sum of the squares of the first number and the last number?

• A. $1201$
• B. $576$
• C. $1236$
• D. $1154$

Answer 1

Answer: (D)

$1154$

Given that, the sum of three consecutive integers is $72$

To find out: The sum of squares of the first and last number.

Let the three consecutive numbers be $x,x+1$, and $x+2$

$\therefore x+(x+1)+(x+2)=72$

$\Rightarrow 3x+3=72$

$\Rightarrow 3x=69$

$\Rightarrow x=\frac{69}{3}=23$

Thus the numbers are $23,24,25$.

$\therefore$ The sum of squares of first and last numbers $={23}^2+{25}^2=529+625=1154$.

Hence, the sum of squares of the first and last number is $1154$.