AS & A Level · Practice questions · Arithmetic progressions

Step by step.

Six original Cambridge-style questions on arithmetic progressions: the \(n\)th term, the sum of \(n\) terms, unknown terms and an AP in context.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Find \(a\) and \(d\), then the right formula.

Identify: \(a\) is the first term, \(d\) the constant step (it can be negative).
One term: \(u_n = a + (n-1)d\).
Sum: \(S_n = \dfrac{n}{2}\left[2a + (n-1)d\right]\) or \(\dfrac{n}{2}(a + l)\).
Unknowns: two facts give two simultaneous equations in \(a\) and \(d\); three terms in AP give \(2b = a + c\).
Inequalities: solve the quadratic, then take the next whole number and check both sides.

Recall

[3 marks]

For the AP \(15, 11, 7, \ldots\) find (a) the 25th term, (b) the sum of the first 25 terms.

\(a = 15\), \(d = -4\). ✓
(a) \(u_{25} = 15 + 24 \times (-4) = -81\) ✓
(b) \(S_{25} = \dfrac{25}{2}(15 + (-81)) = 12.5 \times (-66) = -825\) ✓

Calculation

[4 marks]

The 4th term of an AP is 19 and the sum of the first 10 terms is 235. Find the first term and the common difference.

\(a + 3d = 19\), so \(2a + 6d = 38\). ✓
\(\dfrac{10}{2}(2a + 9d) = 235\), so \(2a + 9d = 47\). ✓
Subtract: \(3d = 9\), \(d = 3\); then \(a = 19 - 9 = 10\). ✓

Proof

[4 marks]

Prove that the sum of the first \(n\) terms of an arithmetic progression with first term \(a\) and common difference \(d\) is \(\dfrac{n}{2}\left[2a + (n-1)d\right]\).

\(S_n = a + (a + d) + \cdots + \left[a + (n-1)d\right]\) ✓
Reversed: \(S_n = \left[a + (n-1)d\right] + \left[a + (n-2)d\right] + \cdots + a\) ✓
Adding term by term, each of the \(n\) pairs sums to \(2a + (n-1)d\), so \(2S_n = n\left[2a + (n-1)d\right]\). ✓
Hence \(S_n = \dfrac{n}{2}\left[2a + (n-1)d\right]\). ✓

Calculation

[4 marks]

The numbers \(k\), \(3k - 2\) and \(4k + 3\) are the first three terms of an arithmetic progression.

(a) Find \(k\) and the common difference.

(b) Find the sum of the first 15 terms.

(a) \(2(3k - 2) = k + 4k + 3\), so \(6k - 4 = 5k + 3\) and \(k = 7\). ✓
The terms are \(7, 19, 31\), so \(d = 12\). ✓
(b) \(S_{15} = \dfrac{15}{2}\left[14 + 14 \times 12\right] = 7.5 \times 182 = 1365\) ✓

Analysis

[4 marks]

The sum of the first \(n\) terms of a progression is \(S_n = 4n^2 - n\).

(a) Find the first two terms.

(b) Show that the progression is arithmetic and find an expression for the \(n\)th term.

(a) \(u_1 = S_1 = 3\); \(S_2 = 14\), so \(u_2 = 11\). ✓
(b) \(u_n = S_n - S_{n-1} = (4n^2 - n) - \left[4(n-1)^2 - (n-1)\right] = 4n^2 - n - 4n^2 + 8n - 4 + n - 1 = 8n - 5\). ✓
\(u_n - u_{n-1} = 8\), a constant, so the progression is arithmetic with \(d = 8\) and \(u_n = 8n - 5\). ✓

Analysis

[6 marks]

A runner trains every day. She runs 2 km on day 1 and increases her distance by 0.25 km each day.

(a) Find the distance she runs on day 20.

(b) Find the total distance she runs in the first 20 days.

(a) \(u_{20} = 2 + 19 \times 0.25 = 6.75\) km ✓
(b) \(S_{20} = \dfrac{20}{2}(2 + 6.75) = 87.5\) km ✓
(c) \(S_n = \dfrac{n}{2}\left[4 + 0.25(n-1)\right] = \dfrac{n(0.25n + 3.75)}{2} > 200\), so \(n^2 + 15n - 1600 > 0\). ✓
Positive root \(n = \dfrac{-15 + \sqrt{6625}}{2} = 33.2\), so \(n = 34\). ✓
Check: \(S_{33} = 198\) km and \(S_{34} = 208.25\) km, so day 34. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.