The sum of 40 consecutive integers is 100. What is the smallest of these 40 integers?

The sum of 40 consecutive integers is 100. What is the smallest of these 40 integers?

$$\small{\text{The sum of a arithmetic sequence is $
\boxed{~~S_n = t_1\cdot \binom{n}{1}+d\cdot \binom{n}{2}~~}
$}}\\\\
\small{\text{We have the sum of 40 consecutive integers is 100 so $d=1$ and $ n = 40 $ and $S_{40}=100$ we have :
}}\\
\small{\text{$
\begin{array}{rcl}
S_{40} = 100 &=& t_1\cdot \binom{40}{1} + \binom{40}{2}\\\\
t_1\cdot \binom{40}{1} &=& 100 - \binom{40}{2}\\\\
t_1 &=& \dfrac{100 - \binom{40}{2} }{\binom{40}{1}} \qquad | \qquad \binom{40}{2} = 39\cdot 20 = 780 \quad \binom{40}{1} = 40 \\\\
t_1 &=& \dfrac{100-780}{40} \\\\
t_1 &=& \dfrac{-680}{40}\\\\
t_1 &=& \dfrac{-68}{4}\\\\
t_1 &=& -\dfrac{68}{4}\\\\
\mathbf{t_1} &\mathbf{=}& \mathbf{-17}
\end{array}
$}}$$

The smallest of these 40 integers is -17

Let n be the first integer...and we have

n + (n + 1) + (n +2) +  ..... + (n + 38) + (n +39)

40n + [(39)(40)/2]  = 100

40n + 780  = 100    subtract 780 from both sides

40n = -680    divide both sides by 40

n = -17