Number and Letter Series

A sequence of terms is given, and the task is to find the next term, a missing term, the term that breaks the pattern, or the sum of many terms. Two methods apply. The first is algebraic: the arithmetic and geometric progressions and the standard sums, where the answer follows from a formula. The second is pattern finding: recognising, in a fixed order, which of a small family of rules generated the terms.

Both are treated: the formulas with their derivations, then a fixed search order for unknown patterns (differences, ratios, known sequences, interleaving, operations on the previous term), and the same ideas applied to letter series by turning letters into their alphabet positions.

In the AP and GP formulas, $a=t_1$ is the first term.

Position convention. Terms are numbered from 1, so $t_1$ is the first term shown. Letters are numbered A = 1 to Z = 26, and counting past Z wraps to A.

Standard sums.

Try these in order.

AP sum. Writing $S_n$ forwards and backwards and adding, each of the $n$ column sums equals $t_1 + t_n$, so $S_n = \frac{n}{2}(t_1 + t_n) = \frac{n}{2}\left(2a + (n-1)d\right)$.

GP sum. $S_n - rS_n = a - ar^n$, so for $r \ne 1$, $S_n = a\frac{r^n - 1}{r - 1}$. For $|r| < 1$, $r^n \to 0$ and the infinite sum is $\frac{a}{1 - r}$.

Why differences work. If $t_n$ is a polynomial of degree $k$ in $n$, its $k$-th differences are constant. Constant first differences give degree at most 1 if the rule holds throughout. Constant nonzero second differences give exact degree 2; zero second differences give degree at most 1. A finite prefix only suggests a continuation, rather than proving it. Under the chosen polynomial rule, extend the bottom row by its constant and add back up.

Telescoping. $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$, so $\sum_{k=1}^{n} \frac{1}{k(k+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}$: all middle terms cancel.

Example 1 (AP). For $7, 11, 15, \ldots$: $a = 7$, $d = 4$. The 25th term is $7 + 24 \times 4 = 103$, and the sum of the first 25 terms is $\frac{25}{2}(7 + 103) = 1375$.

Example 2 (GP). For $3, 6, 12, \ldots$: $a = 3$, $r = 2$. The sum of the first 8 terms is $3 \times \frac{2^8 - 1}{2 - 1} = 3 \times 255 = 765$.

Example 3 (infinite GP). $18 + 6 + 2 + \cdots$ has $r = 1/3$, so the sum is $\frac{18}{1 - 1/3} = 27$.

Example 4 (known sequence). $2, 5, 10, 17, 26, ?$. First differences $3, 5, 7, 9$ (odd numbers) point to squares: the terms are $n^2 + 1$. Next: $6^2 + 1 = 37$.

Example 5 (second differences). $1, 4, 11, 22, 37, ?$.

Example 6 (operation on the previous term). $3, 7, 15, 31, ?$. Each term is twice the previous plus 1 ($3 \to 7 \to 15 \to 31$), so the next is $63$. Equivalently $t_n = 2^{n+1} - 1$.

Example 7 (interleaved). $1, 10, 3, 9, 5, 8, 7, ?$. Odd positions: $1, 3, 5, 7$. Even positions: $10, 9, 8$, so the missing 8th term is $7$.

Example 8 (wrong term). $2, 6, 12, 20, 31, 42$. The rule $n(n+1)$ gives $2, 6, 12, 20, 30, 42$, so 31 is wrong and should be 30.

Example 9 (Fibonacci-type). $2, 3, 5, 8, 13, ?$: each term is the sum of the previous two, so the next is $8 + 13 = 21$.

Example 10 (letter step). B, E, H, K, ?: positions $2, 5, 8, 11$, step 3, so the next is 14 = N.

Example 11 (letter wrap). X, A, D, G, ?: positions $24, 1, 4, 7$ with step 3 modulo 26 ($24 + 3 = 27 \to 1$). The next is 10 = J.

Example 12 (letter pairs). AZ, BY, CX, ?: the first letter moves forward one place and the second backward one place; the pair positions always sum to 27. Next: DW.

Example 13 (telescoping). $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots + \frac{1}{20 \times 21} = 1 - \frac{1}{21} = \frac{20}{21}$.

Example 14 (standard sums). $1^2 + 2^2 + \cdots + 10^2 = \frac{10 \times 11 \times 21}{6} = 385$, and $1^3 + \cdots + 10^3 = 55^2 = 3025$.

Example 15 (missing term in the middle). $3, 8, ?, 24, 35, 48$. Differences of the known neighbours ($8 - 3 = 5$; $35 - 24 = 11$; $48 - 35 = 13$) are consecutive odd numbers, so the full difference row is $5, 7, 9, 11, 13$. The missing term is $8 + 7 = 15$, and $15 + 9 = 24$ confirms it. The terms are $n^2 - 1$ for $n = 2, \ldots, 7$.

Example 16 (differences form a GP). $2, 3, 5, 9, 17, ?$. Differences $1, 2, 4, 8$ double each time, so the next difference is 16 and the next term is $33$. Equivalently $t_n = 2^{n-1} + 1$.

Example 17 (difference of differences). $2, 10, 30, 68, ?$.

Example 18 (mixed AP and GP). $5, 8, 16, 19, 38, ?$. The steps alternate: $+3$, $\times 2$, $+3$, $\times 2$. The next step is $+3$, giving $41$; the one after is $\times 2$, giving $82$.

Example 19 (alphanumeric). A1, C4, E9, G16, ?. Letters skip one (positions $1, 3, 5, 7$, so next 9 = I); numbers are squares $1, 4, 9, 16$, so next 25. Answer: I25. Treat the letter part and the number part as two separate series.

Example 20 (number fixes the letter position). 2B, 4D, 8H, 16P, ?. Each letter is the letter at the position given by the number (B = 2, D = 4, H = 8, P = 16). The numbers double, so next comes 32, and position 32 wraps to $32 - 26 = 6$ = F. Answer: 32F. Forgetting the wrap leaves no letter at all.

Example 21 (letter groups by position). ABD, BDH, CFL, ?. In positions: $(1, 2, 4)$, $(2, 4, 8)$, $(3, 6, 12)$: each group is $(k, 2k, 4k)$. Next: $(4, 8, 16)$ = DHP.

Example 22 (wrong term in a GP). $4, 12, 36, 108, 320, 972$. Ratios are 3 except around 320: $108 \times 3 = 324$ and $324 \times 3 = 972$. So 320 is wrong and should be 324. Note that two ratios ($320/108$ and $972/320$) break, but changing one term repairs both.

Example 23 (factorial and product series). $1, 2, 6, 24, 120, ?$. Ratios $2, 3, 4, 5$ increase by 1, so the next ratio is 6 and the next term is $720 = 6!$.