Mixed AP constraints

Working with arithmetic sequences is a core skill in the HSC Sequences topic. In this problem we practise using two given terms to set up simultaneous equations for the first term $a$ and common difference $d$, then apply the sum formula to find $S_{25}$. Even though the steps are straightforward, seeing a clear, systematic approach will help you handle any “mixed AP constraints” that come up in exams.

Problem Statement

An arithmetic sequence has $a_3=11$ and $a_{12}=47$. Find $a_1$ and $d$, then determine $S_{25}$.

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Solutions

We start by using the general term formula for an arithmetic sequence, $a_n = a + (n-1)d$. Substituting the two given terms gives us a pair of linear equations:

$$a+2d=11,\qquad a+11d=47.$$

To solve for $d$, we subtract the first equation from the second. This eliminates $a$ immediately (the coefficients of $a$ in both equations are 1, so they cancel out):

$$9d=36 \implies d=4,$$

then substituting $d=4$ back into $a+2d=11$ gives $a=11-8=3$.

Now that we have $a_1=3$ and $d=4$, we can find the 25th term:

$$a_{25}=3+24\cdot 4=99,$$

so the sum of the first 25 terms, using $S_n=\frac{n}{2}(a_1+a_n)$, is

$$S_{25}=\frac{25}{2}(3+99)=\frac{25}{2}\cdot 102=1275.$$

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Takeaways

• When a problem gives you two separate terms of an arithmetic sequence, set up the general term equation for each and solve the resulting system for $a$ and $d$.
• Subtracting one equation from the other is usually the quickest way to eliminate $a$ and find $d$.
• Always compute $a_n$ before using $S_n = \frac{n}{2}(a_1+a_n)$; if $a_n$ is not given, find it first with $a_n = a + (n-1)d$.

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Further Readings

HSC Sequences, HSC Mechanics, HSC Collections, HSC Trigonometry