Respuesta :

We can use the general term of the arithmetic series. Look:


[tex]\mathsf{a_n=a_1+(n-1)\cdot r}[/tex]


Where:


[tex]\mathsf{a_1}[/tex]: first term, 15.

[tex]\mathsf{a_n}[/tex]: last term, 87.

n: number of terms (what we looking for);

r: reason, which is equal to the difference between two terms, that is, 3.


Let's go to the calculations.


[tex]\mathsf{a_n=a_1+(n-1)\cdot r}\\\\ \mathsf{87=15+(n-1)\cdot3}\\\\ \mathsf{87-15=(3n-3)}\\\\ \mathsf{72=3n-3}\\\\ \mathsf{72+3=3n}\\\\ \mathsf{75=3n}\\\\ \mathsf{n=\dfrac{75}{3}=25}[/tex]


There are 25 terms.

remotecontrolbrain

Answer:

The series has 25 terms.

Step-by-step explanation:

You could just diligently count, but it looks much better if you use the formula for the n-th term of an arithmetic series like this:

[tex]x_k = 15 + 3k\,\,,k=0,1,...[/tex]

so the k=0 term is 15,  k=1 is 18, etc. Now we solve an equation as follows:

[tex]15+3k=87\implies k = (87-15)/3=24[/tex]

so the last term (87) occurs for k=24, which means, including the 0-th term, there are 25 terms in this series.

The advantage of this method is that it would work for very very long series just as nicely, like 15+18+21+...+ 3015, 3018 (where the diligent method would just be way too boring)