Question

There are three consecutive numbers whose sum is 129. Find the square of the middle one.$\left[3Marks\right]$

Answer 1

Let numbers be x, x + 1, x + 2
According to question,
(x) + (x + 1) + (x + 2) = 129$\left[1Mark\right]$
$\Rightarrow$ (3x+3) = 129
$\Rightarrow$ 3x = 129-3
$\Rightarrow$ 3x = 126
$\Rightarrow$ x = 42$\left[1Mark\right]$
$\Rightarrow$ x+1 = 43
$\therefore$ (x+1)${}^2$ = 43 $\times$ 43 = 1849.$\left[1Mark\right]$