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
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: $15, 18, 21, \ldots, 57$.
To find the number of terms in an arithmetic progression, we can use the formula:
Number of terms = (last term - first term) / common difference + 1
In this case, the first term is 15, the last term is 57, and the common difference is 3. Plugging these values into the formula:
Number of terms = (57 - 15) / 3 + 1
Number of terms = 42 / 3 + 1
Number of terms = 14 + 1
Number of terms = 15
Therefore, the arithmetic progression has 15 terms.