How many terms has the arithmetic progression whose first term is 15 and the last term is 57 given that the common difference is 3

Answer 1

We have $a_1 = 15$, $a_n = 57$, and $d=3$. The $n$th term of an arithmetic sequence can be expressed as $a_n = a_1 + (n-1)d$, so we can plug in our values and solve for $n$: \begin{align*}
57 &= 15 + (n-1)(3)\\
42 &= 3(n-1)\\
14 &= n-1\\
n &= \boxed{15}.
\end{align*} The arithmetic progression has 15 terms: $15, 18, 21, \ldots, 57$.