Arithmetic Progression — chapter notes

• A sequence is a list of numbers in a definite order: $T_1, T_2, T_3, \ldots$ ($T_n$ is the $n$th term).
• An arithmetic progression (AP) is a sequence in which each term after the first is obtained by adding a fixed number $d$, the common difference, to the term before it.
• $d = T_2 - T_1 = T_3 - T_2 = \cdots$: always a term minus the one before it.
• $d > 0$: rising terms; $d < 0$: falling terms; $d = 0$: all terms equal.
• To test a list, work out every difference; one equal pair is not enough.

Example: $-1.5, -4, -6.5, -9$ is an AP with $d = -2.5$; $2, 6, 18, 54$ is not an AP (differences $4, 12, 36$).

Starting from the first term $a$, you add $d$ once for each step; the $n$th term is $n - 1$ steps after the first:

• $T_n$ is a linear expression in $n$: $T_n = dn + (a - d)$. Conversely, if $T_n = pn + q$, the sequence is an AP with $d = p$.
• Some books write $t_n$ or $a_n$ for the $n$th term.

Example: for $13, 8, 3, \ldots$: $T_n = 13 + (n - 1) \times (-5) = 18 - 5n$, so $T_{18} = -72$.

• Is $v$ a term? Solve $a + (n - 1)d = v$. It is a term only if $n$ is a positive whole number.
• Number of terms of a finite AP $a, \ldots, l$: solve $a + (n - 1)d = l$, i.e. $n = \dfrac{l - a}{d} + 1$.
• From the end: read backwards, the AP starts at $l$ with common difference $-d$, so the $m$th term from the end is $l - (m - 1)d$; it is also the $(n - m + 1)$th term from the start.

Example: $9, 16, \ldots, 352$ has $n = 50$ terms; the $10$th term from the end is $352 - 9 \times 7 = 289$, the $41$st term.

• Two terms given ($T_p$ and $T_q$): $T_q - T_p = (q - p)d$ gives $d$; then $a = T_p - (p - 1)d$.
• Two other conditions (sums of terms, a term and a sum): write each in $a$ and $d$ and solve the two linear equations.
• Three unknown terms in AP: take $a - d$, $a$, $a + d$ (sum $3a$).
• Condition for three terms: $a$, $b$, $c$ are in AP exactly when $2b = a + c$ ($b$ is the average of $a$ and $c$).

Example: $T_5 = 31$, $T_{14} = -5$: $9d = -36$, $d = -4$, $a = 47$.

Writing $S_n$ forwards and backwards and adding gives $n$ columns, each equal to $a + l$:

• Use $\dfrac{n}{2}[2a + (n - 1)d]$ when $a$, $d$, $n$ are known; use $\dfrac{n}{2}(a + l)$ when the first and last terms are known.
• If only $a$, $d$ and $l$ are known, find $n$ first.
• Sum of a block of terms: $T_{m+1} + \cdots + T_n = S_n - S_m$.

Example: $-8 - 3 + 2 + \cdots + 112$: $n = 25$, so the sum is $\dfrac{25}{2}(-8 + 112) = 1300$.

• n from a given sum: $\dfrac{n}{2}[2a + (n - 1)d] = S$ is a quadratic in $n$. Solve it and keep only positive whole-number roots; if both roots are valid, check the terms (a root may need a zero or negative "term", which a real problem may not allow).
• Terms from a rule for $S_n$: $T_1 = S_1$, and $T_n = S_n - S_{n-1}$ for $n \ge 2$.

Example: $25, 22, 19, \ldots$ with sum $116$: $3n^2 - 53n + 232 = 0$, so $n = 8$ (the root $\dfrac{29}{3}$ is rejected).

Quantities that change by a fixed amount each day, month, row or step form an AP: savings that grow by the same amount, instalments that rise or fall by the same amount, heights of stacks, temperatures on a climb, distances in successive seconds.

1. Identify $a$ (the first value), $d$ (the fixed change, with its sign) and what $n$ counts.
2. Decide whether a term (one value) or a sum (a total) is asked.
3. Watch the first term: if the first reading is at time $0$, the reading after $k$ steps is the $(k + 1)$th term.
4. Give the answer in words with units, and reject impossible values.

Example: a loan of ₹$30{,}000$ repaid by instalments $750, 900, 1050, \ldots$ needs $16$ instalments, the last being ₹$3{,}000$.