# How do you write the first six terms of the sequence a_n=6-n?

Suppose that $n$ starts from the value zero hence

$n = 0 , 1 , 2 , 3 , \ldots$

Just replace that value of $n$ in the formula ${a}_{n} = 6 - n$ hence

for $n = 0$ we have ${a}_{0} = 6 - 0 = 6$

for $n = 1$ we have ${a}_{1} = 6 - 1 = 5$

for $n = 2$ we have ${a}_{2} = 6 - 2 = 4$

for $n = 3$ we have ${a}_{3} = 6 - 3 = 3$

for $n = 4$ we have ${a}_{4} = 6 - 4 = 2$

for $n = 5$ we have ${a}_{5} = 6 - 5 = 1$