Sum of an AP

This problem deals with the sum of an arithmetic progression, a fundamental skill in the HSC Mathematics Extension 1 Sequences and Series topic. By working through this example, you’ll see how to quickly identify the key pieces of information—first term, common difference, and number of terms—and apply the sum formula efficiently.

Problem Statement

Find the sum of the first $40$ terms of the arithmetic sequence

$$9,\ 13,\ 17,\ \ldots$$

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Solutions

We first identify the first term $a=9$, common difference $d=13-9=4$, and the number of terms $n=40$. To use the sum formula $S_n = \frac{n}{2}(a + l)$, we need the last term $l$, which will be the 40th term. Using the $n$th term formula $T_n = a + (n-1)d$:

$$l=9+39\cdot 4=165.$$

Now we can calculate the sum $S_{40}$:

$$S_{40}=\frac{40}{2}(9+165)=20\cdot 174=3480.$$

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Takeaways

• Always begin by identifying the first term $a$, the common difference $d$, and the number of terms $n$ (or the last term $l$).
• The $n$th term of an AP is $T_n = a + (n-1)d$; use it to find $l$ if it isn’t given directly.
• For the sum of the first $n$ terms, $S_n = \frac{n}{2}(a + l)$ is often the quickest path when both the first and last terms are known.

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Further Readings

HSC Integrals, HSC Distributions, HSC Trigonometry, HSC Complex Numbers