Arithmetic progressions

A sequence $a_1, a_2, a_3, \ldots$ is an arithmetic progression if consecutive terms differ by the same constant, called the common difference $d$:

where: $a_{n+1}$ and $a_n$ are any two consecutive terms, and $d$ is fixed — the same value no matter which pair of neighbours you pick. If $d$ is positive the sequence increases, if $d$ is negative it decreases, and if $d = 0$ every term is identical. Starting from a first term $a$, the sequence unfolds as $a,\ a+d,\ a+2d,\ a+3d,\ \ldots$ Each new term adds one more copy of $d$ than the last, which gives the general (n-th) term:

where: $a$ is the first term, $d$ is the common difference, and $n$ is the position of the term counting from 1 — so $T_1 = a$, exactly one $d$ has been added by $T_2$, and $(n-1)$ copies of $d$ have accumulated by the $n$-th term. Two terms are enough to pin down an entire A.P. Given any two terms, write each as $a + (n-1)d$ and solve the resulting pair of simultaneous linear equations for $a$ and $d$ — this is the single most exam-relevant skill in the topic, and the worked example below drills it directly. The figure below shows how a small A.P. sits on a number line: equal jumps of size $d$, term after term.

If three numbers $a, b, c$ are in A.P., the middle one $b$ is their arithmetic mean (A.M.):

where: $b$ is exactly the average of its neighbours $a$ and $c$. Extending the idea, to insert $n$ arithmetic means $A_1, A_2, \ldots, A_n$ between two numbers $a$ and $b$ so that $a, A_1, \ldots, A_n, b$ forms an A.P., treat $b$ as the $(n+2)$-th term of the whole progression:

where: $n+1$ is the number of equal gaps between $a$ and $b$ once $n$ means are inserted — one gap more than the number of means, since the means split the interval into $n+1$ pieces.

The 4th term of an A.P. is 18 and the 9th term is 38. Find the first term, the common difference, and the sum of the first 15 terms. The problem gives two terms and asks for the parameters that define the whole sequence, plus a partial sum built from those parameters — a two-stage task that appears in nearly every A.P. exam paper in some form, so it is worth internalising the pattern completely rather than memorising just this instance.

Step 1. Translate both given terms using $T_n = a+(n-1)d$: $T_4 = a+3d = 18$ and $T_9 = a+8d = 38$. These are two linear equations in the two unknowns $a$ and $d$.

Step 2. Subtract the first equation from the second to eliminate $a$: $(a+8d) - (a+3d) = 38-18$, which gives $5d = 20$, so $d = 4$.

Step 3. Substitute $d=4$ back into $a+3d=18$: $a + 12 = 18$, so $a = 6$. Check against the other equation: $T_9 = 6 + 8(4) = 6+32 = 38$, which matches — the values are consistent.

Step 4. Find $S_{15}$ using $S_n = \frac{n}{2}\{2a+(n-1)d\}$ with $n=15$, $a=6$, $d=4$: $S_{15} = \frac{15}{2}\{2(6)+14(4)\} = \frac{15}{2}\{12+56\} = \frac{15}{2}(68) = 15 \times 34$. So the first term is $a = \mathbf{6}$, the common difference is $d = \mathbf{4}$, and $\mathbf{S_{15} = 510}$.

Competitive stretch

Insert 4 arithmetic means between 3 and 23: here $a=3$, $b=23$, $n=4$, so $d = \frac{23-3}{4+1} = \frac{20}{5} = 4$, giving means $7, 11, 15, 19$ and the full progression $3,7,11,15,19,23$ — every gap equal to $4$, confirming the answer.

• Confusing the $n$-th term with the sum. $T_n = a+(n-1)d$ gives one term; $S_n = \frac{n}{2}\{2a+(n-1)d\}$ gives the total of $n$ terms — mixing the two formulas is the most common slip in A.P. problems.
• Off-by-one errors when inserting means. Inserting $n$ means between $a$ and $b$ uses $d = \frac{b-a}{n+1}$, not $\frac{b-a}{n}$, because the $n$ means split the interval into $n+1$ equal gaps.
• Misapplying $T_n = S_n - S_{n-1}$ at $n=1$. This identity recovers a term from a sum formula but only holds for $n \ge 2$; at $n=1$, $T_1 = S_1$ directly.
• Treating $2b=a+c$ as proof of order. The condition just says $b$ is the average of $a$ and $c$; it does not by itself fix which term comes first unless the sequence's order is already known.

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