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The arithmetic series 15+18+21+...+84+87 has how many terms?

2 Answers

2 votes

Answer:

The series has 25 terms.

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:


x_k = 15 + 3k\,\,,k=0,1,...

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


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

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)

User IDanil
by
7.7k points
4 votes

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



\mathsf{a_n=a_1+(n-1)\cdot r}


Where:



\mathsf{a_1}: first term, 15.


\mathsf{a_n}: 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.



\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=(75)/(3)=25}


There are 25 terms.

User JohnnyJS
by
7.5k points

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