8 min readConcept sheetPart I (1 mark) and Part II (2 marks); finding an AP from two terms, or three terms from their sum and product, is Part III (5 marks)

Arithmetic progression: nth term and three terms in AP

In an AP each term is the previous term plus a fixed common difference d. The nth term is tₙ = a + (n − 1)d.

nth termtₙ = a + (n − 1)d, where a is the first term and d the common difference

Common differenced = t₂ − t₁ = tₙ − tₙ₋₁

Number of termsn = (l − a)/d + 1, where l is the last term

Three terms in APa, b, c are in AP if 2b = a + c; take them as a − d, a, a + d

Key ideas

AP (arithmetic progression): a, a + d, a + 2d, a + 3d, ....
If d > 0 the terms increase; if d < 0 they decrease; if d = 0 every term is equal (a constant AP).
Adding or subtracting the same number to every term keeps an AP, with the same d.
Multiplying or dividing every term by the same non-zero number k keeps an AP, with common difference kd or d/k.
For sum and product problems, take three terms as a − d, a, a + d. For four terms, take a − 3d, a − d, a + d, a + 3d (common difference 2d).

Example 1: a term and the position of a term

Problem

For the AP 7, 11, 15, 19, ..., find (i) the 25th term (ii) which term is 203.

Working

a = 7 and d = 11 − 7 = 4.
Formula: tₙ = a + (n − 1)d.
(i) t₂₅ = a + 24d = 7 + 24 × 4 = 7 + 96 = 103.
(ii) Put tₙ = 203: 7 + (n − 1) × 4 = 203.
(n − 1) × 4 = 196, so n − 1 = 49.
n = 50.

Answer

t₂₅ = 103, and 203 is the 50th term.

Example 2: the AP from two of its terms

Problem

The 5th term of an AP is 19 and the 12th term is 47. Find the AP and its general term.

Working

Formula: tₙ = a + (n − 1)d.
t₅ = a + 4d = 19 and t₁₂ = a + 11d = 47.
Subtract: 7d = 28, so d = 4.
a = 19 − 4 × 4 = 3.
tₙ = 3 + (n − 1) × 4 = 4n − 1.
Check: t₁₂ = 4 × 12 − 1 = 47.

Answer

The AP is 3, 7, 11, 15, ..., with tₙ = 4n − 1.

Example 3: three terms from their sum and product

Problem

Three numbers in AP have sum 27 and product 693. Find them.

Working

Let the numbers be a − d, a, a + d.
Sum: 3a = 27, so a = 9.
Product: (9 − d) × 9 × (9 + d) = 693, so 81 − d² = 77.
d² = 4, so d = 2 or d = −2.
d = 2 gives 7, 9, 11; d = −2 gives 11, 9, 7.

Answer

The numbers are 7, 9 and 11.

Example 4: a word problem

Problem

Kavya saves ₹ 200 in January, ₹ 250 in February, ₹ 300 in March, and so on. In which month will she save ₹ 750?

Working

The monthly savings form an AP with a = 200 and d = 50.
Formula: tₙ = a + (n − 1)d.
Put tₙ = 750: 200 + (n − 1) × 50 = 750.
(n − 1) × 50 = 550, so n − 1 = 11.
n = 12, the 12th month.

Answer

She saves ₹ 750 in December, the 12th month.

In real life

Rows of chairs in a wedding pandal often increase by a fixed number each row, so the chairs per row form an AP. A piggy bank that gets the same amount every week also grows as an AP.

Exam angle

2 marks: “Find k if 2k − 1, k + 4 and 3k − 3 are three consecutive terms of an AP.” 2(k + 4) = (2k − 1) + (3k − 3) ⇒ 2k + 8 = 5k − 4 ⇒ 3k = 12 ⇒ k = 4. The terms are 7, 8, 9.

2 marks: “How many terms are there in the AP 12, 17, 22, ..., 237?” a = 12, d = 5, l = 237. n = (237 − 12)/5 + 1 = 45 + 1 = 46.

5 marks: “The sum of three consecutive terms of an AP is 36 and their product is 1428. Find the three terms.” Take a − d, a, a + d. 3a = 36, so a = 12. 12(144 − d²) = 1428 gives 144 − d² = 119, so d² = 25 and d = ±5. The terms are 7, 12, 17 (or 17, 12, 7).

Remember it

Start, then step: tₙ = a + (n − 1)d, because you take n − 1 steps of d to reach term n.