Arithmetic Progressions — chapter notes

An arithmetic progression (AP) is a list of numbers $a_1, a_2, a_3, \ldots$ in which each term after the first is obtained by adding a fixed number $d$, the common difference:

The first term is usually called $a$. The common difference can be positive (rising list), negative (falling list) or zero (constant list). An AP is finite if it has a last term $l$, and infinite otherwise.

Test for an AP: compute every difference of consecutive terms; all must be equal. Simplify surds and fractions first: $\sqrt2, \sqrt8, \sqrt{18}$ is $\sqrt2, 2\sqrt2, 3\sqrt2$, an AP with $d = \sqrt2$.

Three consecutive terms: $p, q, r$ are in AP exactly when $2q = p + r$ (the middle term is the average).

With first term $a$ and common difference $d$:

Plotted against their positions $n = 1, 2, 3, \ldots$, the terms lie on a straight line: rising if $d > 0$, falling if $d < 0$, flat if $d = 0$. Multiplying every term by a number $k$, or adding the same number to every term, gives another AP (with difference $kd$, or $d$).

To go from the first term to the $n$th, add $d$ exactly $n - 1$ times:

Uses:

• Any term: $7, 11, 15, \ldots$ has $a_{25} = 7 + 24 \times 4 = 103$.
• Which term: set $a_n$ equal to the number and solve for $n$. In $84, 79, 74, \ldots$, $-1$ is $a_{18}$. If $n$ is not a natural number, the number is not a term ($-2$ gives $n - 1 = 17.2$).
• How many terms: find the first and last terms that fit, then solve $l = a + (n - 1)d$ (multiples of $11$ between $200$ and $600$: $209$ to $594$, $36$ terms).
• First negative term: solve the inequality $a + (n - 1)d < 0$ and take the least natural $n$.

Reading a finite AP backwards gives an AP with first term $l$ and common difference $-d$, so

If an AP has $N$ terms, the $n$th term from the end is the $(N - n + 1)$th term from the start.

AP from two terms: two conditions give two equations in $a$ and $d$. For $a_4 = 22$ and $a_{11} = 64$: subtracting gives $7d = 42$, so $d = 6$ and $a = 4$. Always check with the condition you did not use last.

Write $S_n$ forwards and backwards and add: each of the $n$ columns gives $2a + (n - 1)d$. Hence

Special cases: $1 + 2 + \cdots + n = \dfrac{n(n + 1)}{2}$; the first $n$ odd numbers add to $n^2$; the first $n$ even numbers add to $n(n + 1)$.

Example: $5 + 9 + 13 + \cdots$ ($30$ terms) $= 15[10 + 116] = 1890$. Common slip: writing $a$ instead of $2a$ in the bracket.

Sum given, $n$ unknown: the formula gives a quadratic in $n$. Keep only natural-number roots. Two natural roots can both be valid: for $30, 26, 22, \ldots$, both $n = 6$ and $n = 10$ give $S_n = 120$, because terms $7$ to $10$ ($6, 2, -2, -6$) add to $0$.

Term from sums: $a_1 = S_1$ and $a_n = S_n - S_{n-1}$ for $n \ge 2$. If $S_n = 2n^2 + 5n$, the terms are $7, 11, 15, \ldots$ ($a_n = 4n + 3$). A sum formula $S_n = pn^2 + qn$ always comes from an AP with $d = 2p$; with an extra constant term, the list is not an AP from the first term.

Plan: identify what changes by a fixed amount → write $a$ and $d$ with units → decide whether one value ($a_n$) or a total ($S_n$) is asked → solve (for $n$ if needed, keeping natural numbers) → answer in words with units.

Typical set-ups:

• Savings: ₹$200$ rising by ₹$50$ a month: ₹$650$ in month $10$; ₹$5700$ in a year.
• Instalments: ₹$1000$ rising by ₹$100$ repays ₹$118000$ in $40$ months, last instalment ₹$4900$.
• Arrangements: $24$ rows from $18$ seats, $3$ more per row: last row $87$, total $1260$.
• Salary scales, fares, simple interest: amounts at the end of each year form an AP (compound interest does not).
• Limited resources: with a fixed number of items, compare consecutive sums to find how many complete rows can be built.