Sequences & Series 3 — Arithmetic

Section 1

What makes a sequence arithmetic?

An arithmetic sequence has one simple rule: to get from any term to the next, you add the same fixed amount.

That fixed amount is called the common difference (Carl calls it the gap) and we write it $d$.

The test for arithmetic:

$T_{n+1} - T_{n} = d = \text{constant}$

If $T_{n+1} - T_{n}$ comes out to a number with no $n$ in it, the sequence is arithmetic. If $n$ is still hanging around, it isn't.

Worked example — a sequence that is arithmetic

Prove that $T_n = 3n + 1$ defines an arithmetic sequence.

Step 1 — write $T_{n+1}$ by replacing $n$ with $(n+1)$.

$T_{n+1} = 3(n+1) + 1$

$= 3n + 3 + 1$

$= 3n + 4$

Step 2 — subtract.

$T_{n+1} - T_n = (3n + 4) - (3n + 1)$

$= 3$

$3$ is a constant — no $n$. So the sequence is arithmetic, with common difference $d = 3$.

Worked example — a sequence that isn't

Investigate whether $T_n = 5n^{2} + 2n + 3$ is arithmetic.

$T_{n+1} = 5(n+1)^{2} + 2(n+1) + 3$

$= 5(n^{2} + 2n + 1) + 2n + 2 + 3$

$= 5n^{2} + 10n + 5 + 2n + 5$

$= 5n^{2} + 12n + 10$

$T_{n+1} - T_n = (5n^{2} + 12n + 10) - (5n^{2} + 2n + 3)$

$= 5n^{2} + 12n + 10 - 5n^{2} - 2n - 3$

$= 10n + 7$

$10n + 7$ still contains $n$, so the gap changes as you move through the sequence. Not arithmetic.

Big-picture rule: $T_n$ linear in $n$ $\Rightarrow$ arithmetic. $T_n$ quadratic (or higher) $\Rightarrow$ not arithmetic.

Try this

Prove that $T_n = 4n - 7$ defines an arithmetic sequence and state $d$.

Write $T_{n+1}$, subtract $T_n$, check the result has no $n$.

$T_{n+1} = 4(n+1) - 7 = 4n + 4 - 7 = 4n - 3$

$T_{n+1} - T_n = (4n - 3) - (4n - 7)$

$= 4n - 3 - 4n + 7$

$= 4$ (constant) ✓

Arithmetic with $d = 4$.

Try this

Investigate whether $T_n = n^{2} + 3n$ is arithmetic.

$T_n$ is quadratic — expect $n$ to survive in the difference.

$T_{n+1} = (n+1)^{2} + 3(n+1)$

$= n^{2} + 2n + 1 + 3n + 3$

$= n^{2} + 5n + 4$

$T_{n+1} - T_n = (n^{2} + 5n + 4) - (n^{2} + 3n)$

$= 2n + 4$

Not a constant.

Not arithmetic.

Try this

Prove that $T_n = 7 - 2n$ is arithmetic and state $d$.

A negative $d$ is fine — it just means the sequence decreases.

$T_{n+1} = 7 - 2(n+1) = 7 - 2n - 2 = 5 - 2n$

$T_{n+1} - T_n = (5 - 2n) - (7 - 2n)$

$= 5 - 2n - 7 + 2n$

$= -2$ (constant) ✓

Arithmetic with $d = -2$.

Section 2

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

For arithmetic sequences we have two key labels.

$a$ = first term: $a = T_1 = S_1$

$d$ = common difference (gap): $d = T_2 - T_1$

Reading $a$ and $d$ off a list

$2,\ 5,\ 8,\ 11,\ 14,\ \dots$

First term: $a = 2$.

Gap: $d = 5 - 2 = 3$.

$15,\ 12,\ 9,\ 6,\ 3,\ \dots$

$a = 15,\ \ d = 12 - 15 = -3$.

From a $T_n$ formula

Given $T_n = 5n - 3$, find $a$ and $d$.

$a = T_1 = 5(1) - 3 = 2$

$T_2 = 5(2) - 3 = 7$

$d = T_2 - T_1 = 7 - 2 = 5$

$a = 2, \ \ d = 5$

From $a$ and $d$ back to the terms

Given $a = \dfrac{2}{3}$ and $d = \dfrac{1}{2}$, find the first four terms.

Fractions with different bottoms are a pain to add. Common-denominator everything first:

$a = \dfrac{2}{3} = \dfrac{4}{6}$ $d = \dfrac{1}{2} = \dfrac{3}{6}$

Now keep adding $\dfrac{3}{6}$:

$\dfrac{4}{6}, \ \dfrac{7}{6}, \ \dfrac{10}{6}, \ \dfrac{13}{6}$

From an $S_n$ formula

Given $S_n = n^{2} + 7n$, find $a$ and $d$.

Use the $S \leftrightarrow T$ machinery from Part 2.

$a = T_1 = S_1 = 1 + 7 = 8$

$S_2 = 2^{2} + 7(2) = 4 + 14 = 18$

$T_2 = S_2 - S_1 = 18 - 8 = 10$

$d = T_2 - T_1 = 10 - 8 = 2$

$a = 8, \ \ d = 2$

Calc shortcut (if you've met differentiation): treat $S_n$ as $y = n^{2} + 7n$, then $\dfrac{d^{2}y}{dn^{2}} = 2$ reads off $d$ directly. For any arithmetic series, the second derivative of $S_n$ is the common difference. Quick sanity check; works because $S_n$ is always a quadratic in $n$ for an AP.

Harder case — find $T_n$ first, then read off $a$ and $d$

Given $S_n = n^{2} + 9n + 1$, find $T_n$, hence $a$ and $d$.

$T_n = S_n - S_{n-1}$

$S_{n-1} = (n-1)^{2} + 9(n-1) + 1$

$= n^{2} - 2n + 1 + 9n - 9 + 1$

$= n^{2} + 7n - 7$

$T_n = (n^{2} + 9n + 1) - (n^{2} + 7n - 7)$

$= n^{2} + 9n + 1 - n^{2} - 7n + 7$

$= 2n + 8$

$T_1 = 2(1) + 8 = 10$

$T_2 = 2(2) + 8 = 12$

$d = T_2 - T_1 = 2$

$T_n = 2n + 8, \ \ a = 10, \ \ d = 2$

Try this

For $T_n = 6n + 1$, find $a$ and $d$.

$a = T_1$. Plug in $n=1$ and $n=2$.

$a = T_1 = 6(1) + 1 = 7$

$T_2 = 6(2) + 1 = 13$

$d = 13 - 7 = 6$

$a = 7, \ \ d = 6$

Try this

$a = \dfrac{1}{4}$ and $d = \dfrac{1}{3}$. Find the first four terms.

Common denominator first. $4$ and $3$ share $12$.

$a = \dfrac{1}{4} = \dfrac{3}{12}$ $d = \dfrac{1}{3} = \dfrac{4}{12}$

Add $\dfrac{4}{12}$ each step:

$\dfrac{3}{12}, \ \dfrac{7}{12}, \ \dfrac{11}{12}, \ \dfrac{15}{12}$

Try this

Given $S_n = n^{2} + 5n$, find $a$ and $d$ without finding $T_n$.

$a = S_1$. For $d$: get $T_2 = S_2 - S_1$, then $d = T_2 - T_1$.

$a = S_1 = 1 + 5 = 6$

$S_2 = 4 + 10 = 14$

$T_2 = S_2 - S_1 = 14 - 6 = 8$

$d = 8 - 6 = 2$

(Calc check: $\frac{d^{2}}{dn^{2}}(n^{2}+5n) = 2$ ✓)

$a = 6, \ \ d = 2$

Try this

Given $S_n = 2n^{2} + 3n$, find $T_n$, hence $a$ and $d$.

$T_n = S_n - S_{n-1}$. Watch the $2$ multiplying $(n-1)^{2}$.

$S_{n-1} = 2(n-1)^{2} + 3(n-1)$

$= 2(n^{2} - 2n + 1) + 3n - 3$

$= 2n^{2} - 4n + 2 + 3n - 3$

$= 2n^{2} - n - 1$

$T_n = (2n^{2} + 3n) - (2n^{2} - n - 1)$

$= 4n + 1$

$a = T_1 = 5, \ \ T_2 = 9, \ \ d = 4$

$T_n = 4n + 1, \ \ a = 5, \ \ d = 4$

Section 3

The general term: $T_n = a + (n-1)d$

Write the terms of an arithmetic sequence using $a$ and $d$:

$T_1 = a$

$T_2 = a + d$

$T_3 = a + 2d$

$T_4 = a + 3d$

$\vdots$

Spot the pattern: term number $n$ has $(n-1)$ jumps of $d$ added to $a$.

The general term:

$T_n = a + (n-1)d$

This is in your Tables — you don't need to memorise it.

Worked example (i)

Find $T_n$ for the sequence $3, \ 11, \ 19, \ 27, \ \dots$

$a = 3, \ \ d = 11 - 3 = 8$

$T_n = a + (n-1)d$

$= 3 + 8(n - 1)$

$= 3 + 8n - 8$

$T_n = 8n - 5$

Worked example (ii) — negative $d$

Find $T_n$ for $-3, \ -9, \ -15, \ -21, \ \dots$

$a = -3, \ \ d = -9 - (-3) = -6$

$T_n = -3 + (-6)(n - 1)$

$= -3 - 6n + 6$

$T_n = 3 - 6n$

Try this

Find $T_n$ for $5, \ 9, \ 13, \ 17, \ \dots$

$a$, $d$, plug into $T_n = a + (n-1)d$, expand.

$a = 5, \ \ d = 4$

$T_n = 5 + 4(n-1)$

$= 5 + 4n - 4$

$T_n = 4n + 1$

Try this

Find $T_n$ for $20, \ 17, \ 14, \ 11, \ \dots$

$d$ is negative — watch the sign when expanding.

$a = 20, \ \ d = -3$

$T_n = 20 + (-3)(n-1)$

$= 20 - 3n + 3$

$T_n = 23 - 3n$

Try this

Find $T_n$ for $-7, \ -2, \ 3, \ 8, \ \dots$ and use your formula to find $T_{20}$.

Two steps: get $T_n$, then sub $n=20$.

$a = -7, \ \ d = -2 - (-7) = 5$

$T_n = -7 + 5(n-1)$

$= -7 + 5n - 5$

$= 5n - 12$

$T_{20} = 5(20) - 12 = 100 - 12 = 88$

$T_n = 5n - 12, \ \ T_{20} = 88$

Section 4

The series sum: $S_n = \dfrac{n}{2}\bigl\{2a + (n-1)d\bigr\}$

Adding 40 terms by hand is no good to anyone. There's a closed-form sum, and it comes from a beautiful trick.

The reverse-and-add derivation

$S_n = a + (a+d) + (a+2d) + \dots + (a+(n-2)d) + (a+(n-1)d)$

Write it again with the order reversed:

$S_n = (a+(n-1)d) + (a+(n-2)d) + \dots + (a+d) + a$

Add the two lines vertically. Each column gives the same total: $2a + (n-1)d$. And there are $n$ columns.

$2S_n = n\bigl\{2a + (n-1)d\bigr\}$

Sum of an AP (in Tables):

$S_n = \dfrac{n}{2}\bigl\{2a + (n-1)d\bigr\}$

Worked example

For the AP $2 + 5 + 8 + 11 + \dots$ find $S_n$ and hence $S_{40}$.

$a = 2, \ \ d = 3$

$S_n = \dfrac{n}{2}\bigl\{2(2) + 3(n-1)\bigr\}$

$= \dfrac{n}{2}\bigl\{4 + 3n - 3\bigr\}$

$= \dfrac{n}{2}(3n + 1)$

$S_n = \dfrac{3n^{2} + n}{2}$

Now $S_{40}$:

$S_{40} = \dfrac{40}{2}\bigl\{3(40) + 1\bigr\}$

$= 20 \times 121$

$S_{40} = 2420$

Try this

Find $S_{20}$ for the AP $1 + 4 + 7 + 10 + \dots$

$a$, $d$, then plug into the sum formula with $n=20$.

$a = 1, \ \ d = 3, \ \ n = 20$

$S_{20} = \dfrac{20}{2}\bigl\{2(1) + 3(19)\bigr\}$

$= 10\bigl\{2 + 57\bigr\}$

$= 10 \times 59$

$S_{20} = 590$

Try this

For the AP $30 + 27 + 24 + 21 + \dots$ find $S_{15}$.

$d$ is negative.

$a = 30, \ \ d = -3, \ \ n = 15$

$S_{15} = \dfrac{15}{2}\bigl\{2(30) + (-3)(14)\bigr\}$

$= \dfrac{15}{2}\bigl\{60 - 42\bigr\}$

$= \dfrac{15}{2}(18)$

$= 15 \times 9$

$S_{15} = 135$

Try this

An AP has $a = 7$ and $d = 4$. Find a formula for $S_n$ in simplified form.

Tidy the brace before multiplying by $\dfrac{n}{2}$.

$S_n = \dfrac{n}{2}\bigl\{2(7) + 4(n-1)\bigr\}$

$= \dfrac{n}{2}\bigl\{14 + 4n - 4\bigr\}$

$= \dfrac{n}{2}(4n + 10)$

$= \dfrac{n \cdot 2(2n + 5)}{2}$

$S_n = n(2n + 5)$

Section 5

Sim Eqs — two unknowns, two facts

To pin down an arithmetic sequence you need two numbers: $a$ and $d$. So if a question gives you two facts, set up two equations and solve simultaneously.

Worked example 1 — two known terms

In an arithmetic sequence the $3^{\text{rd}}$ term is $21$ and the $8^{\text{th}}$ term is $53$. Find $a$ and $d$.

Use $T_n = a + (n-1)d$ on each fact:

$T_3 = 21 \ \Rightarrow \ a + 2d = 21$

$T_8 = 53 \ \Rightarrow \ a + 7d = 53$

Subtract the first from the second (the $a$'s cancel):

$5d = 32$

$d = \dfrac{32}{5}$

Back-substitute into $a + 2d = 21$:

$a + 2 \cdot \dfrac{32}{5} = 21$

$a + \dfrac{64}{5} = \dfrac{105}{5}$

$a = \dfrac{41}{5}$

$a = \dfrac{41}{5}, \ \ d = \dfrac{32}{5}$

Worked example 2 — a term and a sum

In an AP the third term is $15$ and the sum of the first $6$ terms is $30$. Find $a$ and $d$.

$T_3 = 15 \ \Rightarrow \ a + 2d = 15$ (1)

$S_6 = 30 \ \Rightarrow \ \dfrac{6}{2}\bigl\{2a + 5d\bigr\} = 30$

$3(2a + 5d) = 30$

$2a + 5d = 10$ (2)

Double (1): $2a + 4d = 30$

Subtract from (2): $d = 10 - 30 = -20$

From (1): $a + 2(-20) = 15 \ \Rightarrow \ a = 55$

$a = 55, \ \ d = -20$

Worked example 3 — relationships between terms

In an AP the sum of the $5^{\text{th}}$ and $6^{\text{th}}$ terms is $20$. The $4^{\text{th}}$ term is $3$ times the $2^{\text{nd}}$ term. Find $a$ and $d$.

Express each term using $T_n = a + (n-1)d$:

$T_5 = a + 4d, \ \ T_6 = a + 5d$

$T_2 = a + d, \ \ T_4 = a + 3d$

Translate the two facts:

$T_5 + T_6 = 20: \ (a + 4d) + (a + 5d) = 20$

$2a + 9d = 20$ (1)

$T_4 = 3 T_2: \ a + 3d = 3(a + d)$

$a + 3d = 3a + 3d$

$0 = 2a$

$a = 0$ (2)

Sub $a = 0$ into (1): $9d = 20 \ \Rightarrow \ d = \dfrac{20}{9}$

$a = 0, \ \ d = \dfrac{20}{9}$

The key move every time: write each term as $a + (\text{something})d$, then the given conditions turn into linear equations in $a$ and $d$.

Try this

In an AP, $T_4 = 14$ and $T_9 = 29$. Find $a$ and $d$.

$a + 3d = 14$, $a + 8d = 29$. Subtract.

$T_4 = 14: \ a + 3d = 14$ (1)

$T_9 = 29: \ a + 8d = 29$ (2)

(2) $-$ (1): $5d = 15 \ \Rightarrow \ d = 3$

Sub into (1): $a + 9 = 14 \ \Rightarrow \ a = 5$

$a = 5, \ \ d = 3$

Try this

In an AP, $T_5 = 11$ and $S_8 = 60$. Find $a$ and $d$.

$T_5$ gives one equation. $S_8$ via $\dfrac{n}{2}\{2a + (n-1)d\}$ gives another.

$T_5 = 11: \ a + 4d = 11$ (1)

$S_8 = 60: \ \dfrac{8}{2}\bigl\{2a + 7d\bigr\} = 60$

$4(2a + 7d) = 60$

$2a + 7d = 15$ (2)

Double (1): $2a + 8d = 22$

Subtract (2): $d = 7$

From (1): $a + 28 = 11 \ \Rightarrow \ a = -17$

$a = -17, \ \ d = 7$

Try this

In an AP, the $6^{\text{th}}$ term is twice the $3^{\text{rd}}$ term, and the $10^{\text{th}}$ term is $30$. Find $a$ and $d$.

Write each term as $a + (\text{k})d$ and translate the conditions.

$T_6 = 2 T_3: \ a + 5d = 2(a + 2d)$

$a + 5d = 2a + 4d$

$d = a$ (1)

$T_{10} = 30: \ a + 9d = 30$ (2)

Sub $d = a$ into (2): $a + 9a = 30 \ \Rightarrow \ 10a = 30 \ \Rightarrow \ a = 3$

Then $d = 3$.

$a = 3, \ \ d = 3$

Try this

In an AP, $T_2 + T_5 = 24$ and $S_{10} = 145$. Find $a$ and $d$.

$T_2 + T_5 = (a+d) + (a+4d)$. Then $S_{10}$ via the formula.

$T_2 + T_5: \ (a + d) + (a + 4d) = 24$

$2a + 5d = 24$ (1)

$S_{10} = 145: \ \dfrac{10}{2}\bigl\{2a + 9d\bigr\} = 145$

$5(2a + 9d) = 145$

$2a + 9d = 29$ (2)

(2) $-$ (1): $4d = 5 \ \Rightarrow \ d = \dfrac{5}{4}$

From (1): $2a + \dfrac{25}{4} = 24 \ \Rightarrow \ 2a = \dfrac{71}{4} \ \Rightarrow \ a = \dfrac{71}{8}$

$a = \dfrac{71}{8}, \ \ d = \dfrac{5}{4}$

Section 6

The $x$-question

Sometimes you're told three consecutive terms of an AP in terms of $x$, and asked to find $x$. There's one neat fact that cracks every one of these.

$x$-question rule: for any three consecutive terms of an AP,

$T_2 - T_1 = T_3 - T_2$

(The gap from term 1 to 2 equals the gap from term 2 to 3 — because the gap is constant.)

Worked example

Three consecutive terms of an arithmetic progression are $3x - 1, \ x - 5, \ 5x + 2$. Find $x$.

Label the terms:

$T_1 = 3x - 1, \ \ T_2 = x - 5, \ \ T_3 = 5x + 2$

Apply $T_2 - T_1 = T_3 - T_2$:

$(x - 5) - (3x - 1) = (5x + 2) - (x - 5)$

The two big minus signs each apply to everything in the bracket they're in front of. Distribute carefully.

$x - 5 - 3x + 1 = 5x + 2 - x + 5$

$-2x - 4 = 4x + 7$

$-6x = 11$

$x = -\dfrac{11}{6}$

Try this

$x + 2, \ 2x + 1, \ 4x - 3$ are three consecutive terms of an AP. Find $x$.

$T_2 - T_1 = T_3 - T_2$. Brackets matter.

$(2x + 1) - (x + 2) = (4x - 3) - (2x + 1)$

$2x + 1 - x - 2 = 4x - 3 - 2x - 1$

$x - 1 = 2x - 4$

$3 = x$

$x = 3$

Try this

$x - 4, \ 2x + 1, \ 5x - 2$ are three consecutive terms of an AP. Find $x$, and state the three terms.

Find $x$ first. Then plug back in to get the three numbers.

$(2x + 1) - (x - 4) = (5x - 2) - (2x + 1)$

$2x + 1 - x + 4 = 5x - 2 - 2x - 1$

$x + 5 = 3x - 3$

$8 = 2x \ \Rightarrow \ x = 4$

Terms: $x - 4 = 0, \ \ 2x + 1 = 9, \ \ 5x - 2 = 18$.

Check: gaps are $9$ and $9$ ✓

$x = 4; \ \ 0, \ 9, \ 18$

Try this

$2x, \ x + 3, \ x^{2} - 9$ are three consecutive terms of an AP. Find both possible values of $x$.

$T_3$ is quadratic in $x$ — the gap equation becomes a quadratic.

$(x + 3) - 2x = (x^{2} - 9) - (x + 3)$

$3 - x = x^{2} - x - 12$

$3 = x^{2} - 12$

$x^{2} = 15$

$x = \sqrt{15} \ \text{ or } \ x = -\sqrt{15}$

$x = \pm\sqrt{15}$

Section 7

Symmetric variables — 3 consecutive terms with a sum

If a question asks for three consecutive terms of an AP and one of the facts is their sum, there's a beautiful trick that kills $d$ on the spot.

Instead of calling the terms $a, \ a+d, \ a+2d$, call them:

Symmetric labels:

$x - y, \ \ x, \ \ x + y$

The gap is $y$ either way, so they form an AP. Their sum is $3x$ — the $y$'s cancel automatically.

Worked example

Find three consecutive terms of an AP which add to $12$ and multiply to $48$.

Call the three terms $x - y, \ x, \ x + y$.

Sum $= 12$:

$(x - y) + x + (x + y) = 12$

$3x = 12$

$x = 4$

Product $= 48$:

$(x - y)(x)(x + y) = 48$

$x(x^{2} - y^{2}) = 48$

Sub $x = 4$:

$4(16 - y^{2}) = 48$

$16 - y^{2} = 12$

$y^{2} = 4$

$y = 2 \ \text{ or } \ y = -2$

The three terms are $2, \ 4, \ 6$ (or $6, \ 4, \ 2$ in reverse).

Either value of $y$ gives the same set of three numbers — one ordering reads up, the other reads down. Both are valid APs.

Try this

Three consecutive terms of an AP add to $15$ and multiply to $80$. Find the terms.

Call them $x - y, \ x, \ x + y$. Sum gives $x$; product then gives $y$.

Sum: $3x = 15 \ \Rightarrow \ x = 5$

Product: $x(x^{2} - y^{2}) = 80$

$5(25 - y^{2}) = 80$

$25 - y^{2} = 16$

$y^{2} = 9 \ \Rightarrow \ y = \pm 3$

Terms: $2, \ 5, \ 8$

$2, \ 5, \ 8$

Try this

Three consecutive terms of an AP add to $21$. The sum of their squares is $179$. Find the terms.

$(x-y)^{2} + x^{2} + (x+y)^{2}$ simplifies neatly — most cross terms cancel.

Sum: $3x = 21 \ \Rightarrow \ x = 7$

Sum of squares: $(x-y)^{2} + x^{2} + (x+y)^{2}$

$= (x^{2} - 2xy + y^{2}) + x^{2} + (x^{2} + 2xy + y^{2})$

$= 3x^{2} + 2y^{2}$

$3(49) + 2y^{2} = 179$

$147 + 2y^{2} = 179$

$2y^{2} = 32 \ \Rightarrow \ y^{2} = 16 \ \Rightarrow \ y = \pm 4$

Terms: $3, \ 7, \ 11$

$3, \ 7, \ 11$

Part 3 done.

You now own the arithmetic toolkit: the constant-gap test, the $T_n = a + (n-1)d$ general term, the $S_n = \dfrac{n}{2}\{2a + (n-1)d\}$ sum, simultaneous equations for $a$ and $d$, $x$-questions, and the symmetric-variables trick.

Part 4 picks up the matching toolkit for geometric sequences — constant ratio $r$, $T_n = ar^{n-1}$, the GP sum, and the infinite sum when $|r| < 1$.