7 min readConcept sheetShort calculation; 10-mark salary or instalment problem

Arithmetic progressions (AP)

In an AP each term changes by the same common difference d. The nth term is a + (n − 1)d, and the sum is n/2 × (first + last).

First terma

Common differenced = T₂ − T₁

nth termTₙ = a + (n − 1)d

Sum of n termsSₙ = n/2 [2a + (n − 1)d]

AP formulas

Sequencea, a + d, a + 2d, a + 3d, …

nth termTₙ = a + (n − 1)d

Sum of n termsSₙ = n/2 [2a + (n − 1)d]

Sum using the last term lSₙ = n/2 (a + l)

Sum 1 + 2 + … + nn(n + 1)/2

Common differenced = any term − the term before it

A term and a sum

Problem

For the AP 7, 11, 15, …, find the 20th term and the sum of the first 20 terms.

Working

a = 7, d = 11 − 7 = 4, n = 20.
T20 = 7 + (20 − 1) × 4 = 7 + 76 = 83.
S20 = 20/2 × (7 + 83) = 10 × 90 = 900.
Check with the other formula: 20/2 [2(7) + 19(4)] = 10 × (14 + 76) = 900. ✓

Answer

T20 = 83 and S20 = 900.

A salary scale

Problem

A Sri Lankan company starts a clerk on Rs. 60,000 a month, with an increment of Rs. 2,500 a month added every year. Find (a) the monthly salary in year 10 and (b) the total salary earned over the 10 years.

Working

Monthly salaries by year form an AP: a = 60,000, d = 2,500.
(a) T10 = 60,000 + 9 × 2,500 = 60,000 + 22,500 = 82,500.
(b) Sum of the 10 yearly monthly rates: S10 = 10/2 × (60,000 + 82,500) = 5 × 142,500 = 712,500.
Each rate is paid for 12 months: total = 12 × 712,500 = 8,550,000.

Answer

(a) Rs. 82,500 a month (b) Rs. 8,550,000 over 10 years.

How many terms?

Problem

How many terms of 5, 9, 13, … must be added to give 230?

Working

a = 5, d = 4. Use Sₙ = n/2 [2a + (n − 1)d] = 230.
n/2 [10 + 4(n − 1)] = 230, so n/2 (4n + 6) = 230 and n(2n + 3) = 230.
2n² + 3n − 230 = 0. D = 9 + 1,840 = 1,849 = 43².
n = (−3 + 43) ÷ 4 = 10 (the negative root makes no sense).
Check: S10 = 10/2 (10 + 36) = 5 × 46 = 230. ✓

Answer

10 terms.

In Sri Lankan business

Fixed yearly salary increments in Sri Lankan firms form an AP, and straight-line depreciation of a machine is an AP that falls by the same rupee amount every year.

Exam angle

Short answer: “Find the 12th term of 5, 9, 13, …” 5 + 11 × 4 = 49.

Structured question: “How many terms of 5, 9, 13, … give a sum of 230?” n(2n + 3) = 230 gives 2n² + 3n − 230 = 0, so n = 10.

Essay: “A clerk starts at Rs. 60,000 a month with a Rs. 2,500 yearly increment. Find the year-10 salary, total 10-year pay, and the first year the monthly salary exceeds Rs. 85,000.” Rs. 82,500; Rs. 8,550,000; year 12 (Rs. 87,500).

Remember it

AP = Add. Sum = average of first and last × number of terms: Sₙ = n × (a + l)/2.