Arithmetic Sequences and Series – A Level Maths Year 2

1 Arithmetic Sequences

An arithmetic sequence (or arithmetic progression, AP) goes up or down by the same amount every time. That fixed amount is the common difference $d$, and the first term is $a$.

The $n$th term of an arithmetic sequence

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

The $(n-1)$ is there because to reach the $n$th term you start at $a$ and take $n - 1$ steps of size $d$. As a recurrence relation an AP is simply $u_{n+1} = u_n + d$.

Worked Example 1 — Using the $n$th term

For the sequence $7, 11, 15, 19, \ldots$ find the $20$th term, and find which term equals $199$.

①$a = 7$, $d = 4$, so $u_n = 7 + 4(n-1) = 4n + 3$.

②$u_{20} = 4(20) + 3 = 83$.

③$4n + 3 = 199 \Rightarrow 4n = 196 \Rightarrow n = 49$. It is the $49$th term.

Worked Example 2 — Two terms given

The $5$th term of an arithmetic sequence is $17$ and the $12$th term is $45$. Find $a$ and $d$.

①$a + 4d = 17$ and $a + 11d = 45$.

②Subtract: $7d = 28$, so $d = 4$.

③$a = 17 - 16 = 1$.

From the $5$th to the $12$th term is $7$ steps, so $7d = 45 - 17$ straight away. Counting steps is quicker than solving equations.

Worked Example 3 — Three consecutive terms

$2k$, $5k - 3$ and $6k + 2$ are consecutive terms of an arithmetic sequence. Find $k$ and the three terms.

①The middle term is the mean of its neighbours, so $2(5k - 3) = 2k + (6k + 2)$.

②$10k - 6 = 8k + 2 \Rightarrow 2k = 8 \Rightarrow k = 4$.

③The terms are $8$, $17$, $26$ (common difference $9$ ✓).

2 Arithmetic Series and the Sum Formula

A series is the sum of the terms of a sequence. $S_n$ means the sum of the first $n$ terms.

Sum of an arithmetic series

$S_n = \dfrac{n}{2}\big[2a + (n-1)d\big] \qquad\text{or}\qquad S_n = \dfrac{n}{2}(a + l)$
where $l = a + (n-1)d$ is the last term

You must be able to prove this formula. The trick is to write the sum twice, once forwards and once backwards.

Proof — The sum of an arithmetic series

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

②Reverse the order: $S_n = \big(a + (n-1)d\big) + \big(a + (n-2)d\big) + \cdots + a$

③Add the two lines term by term. Every pair adds to $2a + (n-1)d$, and there are $n$ pairs:

④$2S_n = n\big[2a + (n-1)d\big]$

⑤$S_n = \dfrac{n}{2}\big[2a + (n-1)d\big]$ $\blacksquare$

Since $2a + (n-1)d = a + \big(a + (n-1)d\big) = a + l$, the same proof gives $S_n = \tfrac{n}{2}(a + l)$: the number of terms times the average of the first and last.

Special case: the first $n$ natural numbers

$\displaystyle\sum_{r=1}^{n} r = 1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}$

This is the case $a = 1$, $d = 1$, $l = n$. For example $1 + 2 + \cdots + 100 = \tfrac12(100)(101) = 5050$.

Worked Example 4 — A straightforward sum

Find the sum of the first $30$ terms of $3 + 7 + 11 + \cdots$

①$a = 3$, $d = 4$, $n = 30$.

②$S_{30} = \dfrac{30}{2}\big[6 + 29 \times 4\big] = 15 \times 122 = 1830$.

Worked Example 5 — Using the last term

Find the sum of all the multiples of $3$ between $100$ and $300$.

①The first multiple is $102$ and the last is $297$.

②Number of terms: $\dfrac{297 - 102}{3} + 1 = 65 + 1 = 66$.

③$S = \dfrac{66}{2}(102 + 297) = 33 \times 399 = 13\,167$.

The "$+1$" is the most common error: $\dfrac{297 - 102}{3}$ counts the gaps, and there is always one more term than gaps.

3 Finding the Number of Terms

When the sum is given and $n$ is unknown, the sum formula becomes a quadratic in $n$. Only a positive integer solution makes sense.

Worked Example 6 — An exact sum

An arithmetic series has first term $5$ and common difference $3$. The sum of the first $n$ terms is $1105$. Find $n$.

①$\dfrac{n}{2}\big[10 + 3(n-1)\big] = 1105 \Rightarrow \dfrac{n}{2}(3n + 7) = 1105$

②$3n^2 + 7n - 2210 = 0$

③$n = \dfrac{-7 \pm \sqrt{49 + 26\,520}}{6} = \dfrac{-7 \pm 163}{6}$

④$n = 26$ (the other root, $-\tfrac{85}{3}$, is rejected).

Worked Example 7 — When does the sum first exceed a value?

For $2 + 5 + 8 + \cdots$, find the least $n$ for which $S_n > 1000$.

①$S_n = \dfrac{n}{2}\big[4 + 3(n-1)\big] = \dfrac{n(3n + 1)}{2}$.

②$\dfrac{n(3n+1)}{2} > 1000 \Rightarrow 3n^2 + n - 2000 > 0$.

③Roots of $3n^2 + n - 2000 = 0$: $n = \dfrac{-1 \pm \sqrt{24\,001}}{6}$, positive root $n \approx 25.65$.

④So $n > 25.65$, and the least integer is $n = 26$.

⑤Check: $S_{25} = \tfrac{25 \times 76}{2} = 950$ and $S_{26} = \tfrac{26 \times 79}{2} = 1027$ ✓

Always check the integers either side of the critical value — it is quick and it catches rounding errors.

Worked Example 8 — From $S_n$ back to $u_n$

The sum of the first $n$ terms of a series is $S_n = 3n^2 + 2n$. Find $u_n$ and show the series is arithmetic.

①$u_n = S_n - S_{n-1}$ (the first $n$ terms minus the first $n - 1$).

②$= 3n^2 + 2n - \big[3(n-1)^2 + 2(n-1)\big] = 3n^2 + 2n - 3n^2 + 6n - 3 - 2n + 2$

③$u_n = 6n - 1$. Check: $u_1 = 5 = S_1$ ✓

④$u_{n+1} - u_n = 6$ is constant, so the series is arithmetic with $a = 5$, $d = 6$.

4 Problems in Context

Anything that increases by a fixed amount each step — a salary with a fixed annual rise, rows of seats, a training plan — is an arithmetic model. Decide carefully what $a$, $d$ and $n$ represent.

Worked Example 9 — A salary scale

Sam's starting salary is £$24\,000$ and rises by £$1500$ each year. (a) Find Sam's total earnings over the first $10$ years. (b) In which year does the salary first exceed £$40\,000$?

①(a) $a = 24\,000$, $d = 1500$, $n = 10$.

②$S_{10} = \dfrac{10}{2}\big[48\,000 + 9 \times 1500\big] = 5 \times 61\,500 = $ £$307\,500$.

③(b) $24\,000 + 1500(n - 1) > 40\,000 \Rightarrow n - 1 > 10.67$.

④$n > 11.67$, so the salary first exceeds £$40\,000$ in year $12$ (when it is £$40\,500$).

Read whether the question wants a single year's value ($u_n$) or a running total ($S_n$). Mixing them up is the commonest error in context questions.

5 Quick Reference

$n$th term

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

Sum

$S_n = \tfrac{n}{2}[2a + (n-1)d]$

Sum with last term

$S_n = \tfrac{n}{2}(a + l)$

Natural numbers

$\sum_{r=1}^{n} r = \tfrac12 n(n+1)$

Proof

Write $S_n$ forwards and backwards, then add.

Number of terms

$\dfrac{l - a}{d} + 1$

Consecutive terms

$2 \times \text{middle} = \text{first} + \text{third}$

From $S_n$ to $u_n$

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

Unknown $n$

Quadratic in $n$; keep the positive integer.

6 Practice Questions

Question 1

Find the $25$th term of the arithmetic sequence $5, 9, 13, \ldots$

▶ Show solution

$a = 5$, $d = 4$: $u_{25} = 5 + 24 \times 4 = 101$.

Question 2

Find the sum of the first $20$ terms of $2 + 6 + 10 + \cdots$

▶ Show solution

$S_{20} = \dfrac{20}{2}\big[4 + 19 \times 4\big] = 10 \times 80 = 800$.

Question 3

The $4$th term of an arithmetic sequence is $10$ and the $9$th term is $30$. Find the first term, the common difference, and the sum of the first $15$ terms.

▶ Show solution

Five steps from $u_4$ to $u_9$: $5d = 20$, so $d = 4$.

$a + 3d = 10 \Rightarrow a = -2$.

$S_{15} = \dfrac{15}{2}\big[-4 + 14 \times 4\big] = \dfrac{15}{2} \times 52 = 390$.

Question 4

Evaluate $\displaystyle\sum_{r=1}^{40} (5r - 2)$.

▶ Show solution

This is arithmetic with $a = 3$, $l = 198$, $n = 40$.

$S = \dfrac{40}{2}(3 + 198) = 20 \times 201 = 4020$.

Or: $5 \times \tfrac12(40)(41) - 2 \times 40 = 4100 - 80 = 4020$.

Question 5

$x$, $2x + 3$ and $4x - 1$ are consecutive terms of an arithmetic sequence. Find $x$ and the common difference.

▶ Show solution

$2(2x + 3) = x + (4x - 1) \Rightarrow 4x + 6 = 5x - 1 \Rightarrow x = 7$.

The terms are $7, 17, 27$, so $d = 10$.

Question 6

Find the sum of all the odd numbers between $100$ and $200$.

▶ Show solution

First $101$, last $199$, $d = 2$.

Number of terms: $\dfrac{199 - 101}{2} + 1 = 50$.

$S = \dfrac{50}{2}(101 + 199) = 25 \times 300 = 7500$.

Question 7

Prove that the sum of the first $n$ odd numbers is $n^2$.

▶ Show solution

The odd numbers $1, 3, 5, \ldots$ form an AP with $a = 1$, $d = 2$.

$S_n = \dfrac{n}{2}\big[2 + 2(n-1)\big] = \dfrac{n}{2} \times 2n = n^2$. $\blacksquare$

Question 8

An arithmetic series has first term $30$ and common difference $-4$. (a) Find the value of $n$ for which $S_n = 0$. (b) Find the greatest value of $S_n$.

▶ Show solution

$S_n = \dfrac{n}{2}\big[60 - 4(n-1)\big] = \dfrac{n}{2}(64 - 4n) = n(32 - 2n)$.

(a) $n(32 - 2n) = 0$ gives $n = 16$ (ignoring $n = 0$).

(b) The sum is greatest while the terms are still positive. The terms are $30, 26, \ldots, 6, 2, -2, \ldots$, so the first $8$ terms are positive.

$S_8 = 8(32 - 16) = 128$. (Equivalently, $n(32-2n)$ is a quadratic with its maximum at $n = 8$.)

Question 9

Find the least number of terms of $3 + 7 + 11 + \cdots$ needed for the sum to exceed $500$.

▶ Show solution

$S_n = \dfrac{n}{2}\big[6 + 4(n-1)\big] = n(2n + 1)$.

$2n^2 + n - 500 > 0$. Positive root: $n = \dfrac{-1 + \sqrt{4001}}{4} \approx 15.56$.

So $n = 16$. Check: $S_{15} = 15 \times 31 = 465$, $S_{16} = 16 \times 33 = 528$ ✓

Question 10

Priya starts a training plan. On day $1$ she runs $2$ km, and each day she runs $0.5$ km further than the day before.

(a) How far does she run on day $15$? (b) Find her total distance over the first $15$ days. (c) Show that her total distance after $n$ days is $\dfrac{n(n+7)}{4}$ km, and find the number of days needed for her total to reach $200$ km. (d) Comment on the suitability of the model for large $n$.

▶ Show solution

(a) $u_{15} = 2 + 14 \times 0.5 = 9$ km.

(b) $S_{15} = \dfrac{15}{2}(2 + 9) = 82.5$ km.

(c) $S_n = \dfrac{n}{2}\big[4 + 0.5(n - 1)\big] = \dfrac{n}{2}\cdot\dfrac{n + 7}{2} = \dfrac{n(n+7)}{4}$ ✓

$\dfrac{n(n+7)}{4} \geqslant 200 \Rightarrow n^2 + 7n - 800 \geqslant 0$.

$n = \dfrac{-7 + \sqrt{49 + 3200}}{2} = \dfrac{-7 + 57}{2} = 25$.

So she reaches exactly $200$ km at the end of day $25$ (check: $\tfrac{25 \times 32}{4} = 200$ ✓).

(d) The daily distance grows without limit — on day $81$ she would run $42$ km, a marathon every day. In reality there is a maximum distance she can run, so the model is only sensible for a limited period.

Arithmetic Sequences and Series · A Level Maths Year 2 · Created with MathJax