Chapter 5 — Arithmetic Progressions

Look at these lists of numbers: 1, 2, 3, 4, ...; 100, 70, 40, 10, ...; −3, −2, −1, 0, ...; 3, 3, 3, 3, ...; 1.5, 1.7, 1.9, 2.1, ... In each, every term after the first is obtained by adding a fixed number to the term before it: 1 in the first list, −30 in the second, 1 in the third, 0 in the fourth, 0.2 in the fifth. Such a list is called an arithmetic progression, abbreviated AP. The fixed number added is the common difference, denoted d; it can be positive, negative or zero. The first term is denoted a.

Formally, a sequence a₁, a₂, a₃, ... is an AP if a₂ − a₁ = a₃ − a₂ = a₄ − a₃ = ... = d. So to test whether a list is an AP, subtract each term from the next and see whether the differences are all equal. For 2, 4, 8, 16 the differences are 2, 4, 8, not equal, so it is not an AP (it is a geometric progression, studied later). For 0.2, 0.22, 0.222, ... the differences are 0.02, 0.002, ..., not equal, not an AP. For 1, 3, 9, 27, ... not an AP. For a, 2a, 3a, 4a, ... the difference is a throughout, an AP. For √2, √8, √18, √32, ... write them as √2, 2√2, 3√2, 4√2; the difference is √2, an AP. For 1², 3², 5², 7², ... that is 1, 9, 25, 49, the differences are 8, 16, 24, not an AP. For 1², 5², 7², 73, that is 1, 25, 49, 73, the differences are all 24, so it is an AP.

The general form of an AP with first term a and common difference d is a, a + d, a + 2d, a + 3d, .... Given a and d we can write as many terms as we like: a = 10, d = 10 gives 10, 20, 30, 40; a = −2, d = 0 gives −2, −2, −2, −2; a = 4, d = −3 gives 4, 1, −2, −5; a = −1, d = 1/2 gives −1, −1/2, 0, 1/2; a = −1.25, d = −0.25 gives −1.25, −1.5, −1.75, −2.

An AP may be finite, with a last term, such as the number of students in classes or the rungs of a ladder, or infinite, going on for ever, such as the multiples of 5. The last term of a finite AP is denoted l.

Examples from life: the taxi fare of Rs 15 for the first km and Rs 8 for each further km gives 15, 23, 31, 39, ... an AP with a = 15, d = 8. The amount of air in a cylinder when a pump removes a quarter each time gives V, 3V/4, 9V/16, ... not an AP (each term is three-quarters of the last, not a fixed amount less). The cost of digging a well, Rs 150 for the first metre and rising by Rs 50 per metre, gives 150, 200, 250, ... an AP. Money in a bank at compound interest is not an AP, because the interest grows each year. Deciding which situations are APs, and giving a and d, is the first examination task of this chapter.

The first term a is simply the first number of the list, and the common difference d is any term minus the one before it: d = a₂ − a₁. Take care with signs and fractions: for 3, 1, −1, −3, d = 1 − 3 = −2; for −5, −1, 3, 7, d = −1 − (−5) = 4; for 1/3, 5/3, 9/3, 13/3, d = 4/3; for 0.6, 1.7, 2.8, 3.9, d = 1.1.

Writing an AP when a and d are given is direct. When conditions are given instead, use the general form and solve. The AP whose first term is 5 and whose second term is 8 has d = 3: 5, 8, 11, 14. The AP whose third term is 12 and whose seventh term is 24: a + 2d = 12 and a + 6d = 24; subtracting, 4d = 12, d = 3, a = 6; the AP is 6, 9, 12, 15, ... Such problems become routine once the nth term formula of the next section is available, but the idea, two equations in a and d, is already visible.

Terms in an AP taken together. When three unknown terms of an AP are needed, take them as a − d, a, a + d; their sum is 3a, which removes one unknown at once. For four terms take a − 3d, a − d, a + d, a + 3d, with common difference 2d and sum 4a. For five, a − 2d, a − d, a, a + d, a + 2d.

Worked example. Three numbers in AP have sum 24 and product 440. Let them be a − d, a, a + d. Sum 3a = 24, a = 8. Product (8 − d)(8)(8 + d) = 440, so 64 − d² = 55, d² = 9, d = ±3. The numbers are 5, 8, 11 (or 11, 8, 5).

Worked example. Four numbers in AP have sum 20 and the sum of their squares is 120. Let them be a − 3d, a − d, a + d, a + 3d. Sum 4a = 20, a = 5. Squares: (5 − 3d)² + (5 − d)² + (5 + d)² + (5 + 3d)² = 4(25) + 20d² = 100 + 20d² = 120, so d² = 1, d = ±1. The numbers are 2, 4, 6, 8.

Worked example. The angles of a triangle are in AP and the largest is twice the smallest. Angles a − d, a, a + d; sum 3a = 180, a = 60; a + d = 2(a − d) gives 60 + d = 120 − 2d, 3d = 60, d = 20. Angles 40°, 60°, 80°.

Arithmetic mean. If a, b, c are in AP then b − a = c − b, so b = (a + c)/2: the middle term is the average of its neighbours, and is called the arithmetic mean of a and c. This gives a quick test and a quick way to find a missing term: if 5, x, 11 are in AP, x = 8; if 2k + 1, 3k + 3, 5k − 1 are in AP, then 2(3k + 3) = (2k + 1) + (5k − 1), so 6k + 6 = 7k, k = 6. Find k so that k + 9, 2k − 1 and 2k + 7 are in AP: 2(2k − 1) = k + 9 + 2k + 7, 4k − 2 = 3k + 16, k = 18.

The examination often asks for the next few terms, for a and d, for a missing term, or for numbers in AP satisfying a condition. The tools are the general form, the symmetric choice of terms, and the arithmetic-mean property.

In an AP with first term a and common difference d, the second term is a + d, the third a + 2d, the fourth a + 3d, and so on; each term has one fewer d than its position number. So the nth term, denoted a_n (also t_n), is a_n = a + (n − 1)d. This is the most important formula of the chapter. It is also called the general term because putting n = 1, 2, 3, ... generates every term. The last term of a finite AP of n terms is l = a + (n − 1)d.

Finding a term. Find the 10th term of 2, 7, 12, ... a = 2, d = 5, n = 10: a₁₀ = 2 + 9 × 5 = 47. Find the 30th term of 10, 7, 4, ...: a = 10, d = −3: a₃₀ = 10 + 29(−3) = −77. Find the 11th term of −3, −1/2, 2, ...: d = −1/2 − (−3) = 5/2: a₁₁ = −3 + 10(5/2) = 22. Find the 20th term from the last of 3, 8, 13, ..., 253: reverse the AP, first term 253, d = −5: 253 + 19(−5) = 158.

Finding n (which term?). Which term of 21, 18, 15, ... is −81? a = 21, d = −3: 21 + (n − 1)(−3) = −81, so −3(n − 1) = −102, n − 1 = 34, n = 35. Is 0 a term? 21 − 3(n − 1) = 0 gives n − 1 = 7, n = 8; yes, the 8th term. Which term of 3, 8, 13, ... is 78? 3 + 5(n − 1) = 78, n − 1 = 15, n = 16.

Is a given number a term? Check whether 301 is a term of 5, 11, 17, 23, ...: 5 + 6(n − 1) = 301, 6(n − 1) = 296, n − 1 = 49.33..., not a whole number, so 301 is not a term. Check whether −150 is a term of 11, 8, 5, 2, ...: 11 − 3(n − 1) = −150, 3(n − 1) = 161, n − 1 = 53.67, not a term. The rule: n must come out a positive integer.

Finding the number of terms. How many terms are in 7, 13, 19, ..., 205? 7 + 6(n − 1) = 205, 6(n − 1) = 198, n − 1 = 33, n = 34. How many terms in 18, 15½, 13, ..., −47? a = 18, d = −5/2: 18 − (5/2)(n − 1) = −47, (5/2)(n − 1) = 65, n − 1 = 26, n = 27. How many two-digit numbers are divisible by 3? They are 12, 15, ..., 99: 12 + 3(n − 1) = 99, n − 1 = 29, n = 30. How many multiples of 4 lie between 10 and 250? 12, 16, ..., 248: 12 + 4(n − 1) = 248, n = 60.

Finding a and d from two terms. The 11th term of an AP is 38 and the 16th term is 73. a + 10d = 38, a + 15d = 73; subtract: 5d = 35, d = 7, a = −32. The 31st term is −32 + 30 × 7 = 178. The 3rd term is 4 and the 9th is −8: a + 2d = 4, a + 8d = −8; 6d = −12, d = −2, a = 8. Which term is zero? 8 − 2(n − 1) = 0, n = 5. The 17th term exceeds the 10th by 7: (a + 16d) − (a + 9d) = 7d = 7, d = 1; note that a cannot be found and is not needed.

Missing terms. Fill in: 2, _, 26: the middle term is (2 + 26)/2 = 14. Fill in: _, 13, _, 3: a + d = 13, a + 3d = 3; 2d = −10, d = −5, a = 18; the AP is 18, 13, 8, 3. Fill in: 5, _, _, 9½: a = 5, a + 3d = 9.5, d = 1.5; 5, 6.5, 8, 9.5.

Every one of these problems is the same formula used four ways: given a, d, n find a_n; given a, d, a_n find n; given two terms find a and d; given a, d, l find n. Write the formula, substitute what is known, solve for what is not.

The nth term formula solves many practical questions, provided the situation is first recognised as an AP and a, d and n are correctly identified.

Savings. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his salary reach Rs 7000? The salaries 5000, 5200, 5400, ... form an AP with a = 5000, d = 200. 5000 + 200(n − 1) = 7000, 200(n − 1) = 2000, n − 1 = 10, n = 11. The 11th year of service, that is 2005.

Weekly savings. Ramkali saved Rs 5 in the first week of a year and increased her weekly saving by Rs 1.75 each week. In which week did her weekly saving become Rs 20.75? a = 5, d = 1.75: 5 + 1.75(n − 1) = 20.75, 1.75(n − 1) = 15.75, n − 1 = 9, n = 10. The 10th week.

Depreciation and growth. A machine loses Rs 2000 of value each year from a starting value of Rs 40,000. After how many years is it worth Rs 10,000? The value after k years is 40000 − 2000k, so 2000k = 30000, k = 15 years. As an AP the values 40000, 38000, ... have a = 40000 and d = −2000, and the value after k years is the (k + 1)th term; define the terms clearly to avoid an off-by-one error.

Bacteria or a plant. A plant is 8 cm tall and grows 1.5 cm every day. On which day will it be 50 cm? 8 + 1.5(n − 1) = 50, 1.5(n − 1) = 42, n − 1 = 28; on the 29th day.

Counting numbers with a property. How many three-digit numbers are divisible by 7? They are 105, 112, ..., 994: 105 + 7(n − 1) = 994, n − 1 = 127, n = 128. How many numbers between 101 and 999 are divisible by both 2 and 5, that is by 10? 110, 120, ..., 990: n = (990 − 110)/10 + 1 = 89. Which is the first negative term of 20, 19¼, 18½, ...? a = 20, d = −3/4: 20 − (3/4)(n − 1) < 0 when n − 1 > 80/3 = 26.67, so n − 1 = 27, n = 28. The 28th term, which is 20 − 81/4 = −1/4.

Terms with a given ratio or difference. The 8th term of an AP is zero; show the 38th term is three times the 18th. a + 7d = 0, so a = −7d. a₃₈ = a + 37d = 30d; a₁₈ = a + 17d = 10d; so a₃₈ = 3a₁₈. If 7 times the 7th term equals 11 times the 11th term, show that the 18th term is zero: 7(a + 6d) = 11(a + 10d) gives 7a + 42d = 11a + 110d, −4a = 68d, a = −17d; then a₁₈ = a + 17d = 0. If m times the mth term equals n times the nth term (m ≠ n), then the (m + n)th term is zero, by the same argument.

Two APs with equal terms. For what n are the nth terms of 63, 65, 67, ... and 3, 10, 17, ... equal? 63 + 2(n − 1) = 3 + 7(n − 1), 60 = 5(n − 1), n = 13. Do the APs 3, 8, 13, ... and 6, 11, 16, ... ever have the same nth term? 3 + 5(n − 1) = 6 + 5(n − 1) never; the difference between the two APs is always 3. The difference between the 100th terms of 2, 5, 8, ... and 7, 10, 13, ... is the same as the difference of their first terms, 5, since they have the same d.

In each problem, write one sentence identifying the AP and its a and d, then the formula, then the solution, then the answer in the words of the question; and check the answer by counting back a few terms.

A teacher, wanting some quiet, asked a class to add the numbers from 1 to 100. The ten-year-old Carl Friedrich Gauss wrote 5050 within moments. His idea: write the sum forwards and backwards and add the two rows.

S = 1 + 2 + 3 + ... + 98 + 99 + 100
S = 100 + 99 + 98 + ... + 3 + 2 + 1
2S = 101 + 101 + 101 + ... + 101 + 101 + 101 (100 times)
So 2S = 100 × 101, S = 5050.

The same trick works for any AP. Let S_n = a + (a + d) + (a + 2d) + ... + [a + (n − 1)d]. Write it in reverse: S_n = [a + (n − 1)d] + [a + (n − 2)d] + ... + a. Adding term by term, each pair sums to 2a + (n − 1)d, and there are n pairs: 2S_n = n[2a + (n − 1)d]. Hence

S_n = (n/2)[2a + (n − 1)d]

Since the last term l = a + (n − 1)d, we may write 2a + (n − 1)d = a + l, giving the second form S_n = (n/2)(a + l): the sum is the number of terms times the average of the first and last terms. Use the first form when a, d and n are known; the second when the first and last terms are known.

Worked examples. Sum of the first 22 terms of 8, 3, −2, ...: a = 8, d = −5, n = 22: S = 11[16 + 21(−5)] = 11(16 − 105) = 11(−89) = −979. Sum of 34 + 32 + 30 + ... + 10: a = 34, d = −2, l = 10, n = (10 − 34)/(−2) + 1 = 13: S = (13/2)(34 + 10) = 13 × 22 = 286. Sum of −5 + (−8) + (−11) + ... + (−230): d = −3, n = (−230 + 5)/(−3) + 1 = 76: S = 38(−5 − 230) = 38(−235) = −8930. Sum of the first 1000 positive integers: (1000/2)(1 + 1000) = 500500. Sum of the first n positive integers: n(n + 1)/2. Sum of the first 15 multiples of 8: 8 + 16 + ... + 120 = (15/2)(8 + 120) = 15 × 64 = 960. Sum of the odd numbers between 0 and 50: 1 + 3 + ... + 49, n = 25: (25/2)(1 + 49) = 625. Sum of 7 + 10½ + 14 + ... + 84: d = 7/2, n = (84 − 7)/(7/2) + 1 = 23: S = (23/2)(7 + 84) = (23 × 91)/2 = 1046.5.

Notice that in the first form the sum is a quadratic expression in n: S_n = (d/2)n² + (a − d/2)n. This is why the sums of an AP grow like a parabola, not a line, and it is the basis of the next section's converse.

The formula S_n = (n/2)[2a + (n − 1)d] involves four quantities; given any three, the fourth can be found. Some cases lead to a quadratic in n, and then the positive integer root is chosen.

Finding n from the sum. How many terms of 24, 21, 18, ... must be taken so that their sum is 78? a = 24, d = −3: (n/2)[48 − 3(n − 1)] = 78, so n(51 − 3n) = 156, 3n² − 51n + 156 = 0, n² − 17n + 52 = 0, (n − 4)(n − 13) = 0, n = 4 or 13. Both are valid: the first 4 terms sum to 24 + 21 + 18 + 15 = 78, and the terms from the 5th to the 13th (12, 9, 6, 3, 0, −3, −6, −9, −12) sum to zero, so the first 13 also sum to 78. How many terms of 9, 17, 25, ... give a sum of 636? (n/2)[18 + 8(n − 1)] = 636, n(8n + 10) = 1272, 8n² + 10n − 1272 = 0, 4n² + 5n − 636 = 0, (n − 12)(4n + 53) = 0, n = 12 (reject the negative). How many terms of −6, −11/2, −5, ... give −25? d = 1/2: (n/2)[−12 + (n − 1)/2] = −25, so n[−12 + (n − 1)/2] = −50, so n(n − 25)/2 = −50, so n² − 25n + 100 = 0, (n − 5)(n − 20) = 0, n = 5 or 20. Both valid, since the terms from the 6th to the 20th sum to zero.

Finding d or a from the sum. Given a = 5, l = 45 and S = 400, find n and d: 400 = (n/2)(5 + 45) = 25n, n = 16; 45 = 5 + 15d, d = 8/3. Given a = 8, a_n = 62, S_n = 210, find n and d: 210 = (n/2)(70), n = 6; 62 = 8 + 5d, d = 54/5. Given d = 5, S₉ = 75, find a and a₉: 75 = (9/2)(2a + 40), 2a + 40 = 50/3, a = −35/3; a₉ = −35/3 + 40 = 85/3. Given a = 2, d = 8, S_n = 90, find n and a_n: 90 = (n/2)(4 + 8(n − 1)) = n(4n − 2), 4n² − 2n − 90 = 0, 2n² − n − 45 = 0, (2n + 9)(n − 5) = 0, n = 5; a₅ = 2 + 32 = 34. Given a_n = 4, d = 2, S_n = −14, find n and a: 4 = a + 2(n − 1), so a = 6 − 2n; −14 = (n/2)(a + 4) = (n/2)(10 − 2n) = n(5 − n), so n² − 5n − 14 = 0, (n − 7)(n + 2) = 0, n = 7, a = −8. Given a = 3, n = 8, S = 192, find d: 192 = 4(6 + 7d), 6 + 7d = 48, d = 6. Given l = 28, S = 144 and 9 terms, find a: 144 = (9/2)(a + 28), a + 28 = 32, a = 4.

Using two sums. The sum of the first 7 terms of an AP is 49 and of the first 17 terms is 289. Find the sum of the first n terms. (7/2)(2a + 6d) = 49 gives a + 3d = 7; (17/2)(2a + 16d) = 289 gives a + 8d = 17; subtracting, 5d = 10, d = 2, a = 1. S_n = (n/2)(2 + 2(n − 1)) = n². The AP is 1, 3, 5, ..., the odd numbers, whose sums are the perfect squares. If the sum of the first p terms equals the sum of the first q terms (p ≠ q), the sum of the first p + q terms is zero: from S_p = S_q, (p/2)[2a + (p − 1)d] = (q/2)[2a + (q − 1)d], which simplifies to 2a(p − q) + d(p² − p − q² + q) = 0, then (p − q)[2a + (p + q − 1)d] = 0, so 2a + (p + q − 1)d = 0, hence S_p+q = ((p + q)/2)[2a + (p + q − 1)d] = 0.

Sum of a block of terms. The sum of terms from the (m + 1)th to the nth is S_n − S_m. Sum of the 11th to the 20th terms of 3, 7, 11, ...: S₂₀ − S₁₀ = 10(6 + 76) − 5(6 + 36) = 820 − 210 = 610. Alternatively, the block is itself an AP with first term a₁₁ = 43, ten terms and d = 4: 5(86 + 36) = 610.

These are the four-mark staples of the chapter. Set up the equations from the formulas, solve, and reject a negative or non-integer n. When a quadratic in n gives two positive integer roots, both are correct, and the reason (a block of terms summing to zero) should be stated.

The sum of the first n terms includes the nth term; the sum of the first n − 1 terms does not. So the nth term is the difference: a_n = S_n − S_n−1 for n ≥ 2, and a₁ = S₁. This relation lets us recover the AP when only a formula for its sums is given.

Worked example 1. The sum of the first n terms of an AP is S_n = 4n − n². Find the first term, the sum of the first two terms, the second term, and the 3rd, 10th and nth terms. S₁ = 4 − 1 = 3, so a₁ = 3. S₂ = 8 − 4 = 4, so a₂ = S₂ − S₁ = 1. S₃ = 12 − 9 = 3, so a₃ = 3 − 4 = −1. In general a_n = S_n − S_n−1 = (4n − n²) − [4(n − 1) − (n − 1)²] = 4n − n² − 4n + 4 + n² − 2n + 1 = 5 − 2n. Check: a₁ = 3, a₂ = 1, a₃ = −1, a₁₀ = −15. The AP is 3, 1, −1, −3, ... with d = −2.

Worked example 2. If S_n = 3n² + 5n, find the AP and its 25th term. a_n = (3n² + 5n) − [3(n − 1)² + 5(n − 1)] = 3n² + 5n − 3n² + 6n − 3 − 5n + 5 = 6n + 2. So a₁ = 8, a₂ = 14, d = 6, and a₂₅ = 152. Check by S₁ = 8 and S₂ = 22, so a₂ = 14.

Worked example 3 (the general principle). A sum formula of the form S_n = An² + Bn (with no constant term) always comes from an AP, with a_n = 2An + (B − A), d = 2A and a = A + B. For S_n = 2n² + 3n: d = 4, a = 5, AP 5, 9, 13, ... If the sum formula has a constant term, as in S_n = n² + 1, then a₁ = S₁ = 2 but a_n = 2n − 1 for n ≥ 2 gives a₂ = 3, a₃ = 5, and the sequence 2, 3, 5, 7, ... is not an AP because its first difference is 1 and the others are 2. So a quadratic sum formula with a non-zero constant does not give an AP.

Worked example 4. The sum of the first n terms of an AP is (5n² + 3n)/2. Find its 20th term. a_n = [5n² + 3n − 5(n − 1)² − 3(n − 1)]/2 = [5n² + 3n − 5n² + 10n − 5 − 3n + 3]/2 = (10n − 2)/2 = 5n − 1. a₂₀ = 99. Check: a₁ = 4 = S₁ = (5 + 3)/2.

Converse use. Show that the sequence whose nth term is 3n + 2 is an AP: a_n − a_n−1 = (3n + 2) − (3n − 1) = 3, a constant, so it is an AP with d = 3 and a = 5. Show that a_n = n² + 1 is not an AP: a_n − a_n−1 = 2n − 1, which depends on n. In general, a_n = pn + q (linear in n) always gives an AP with d = p and first term p + q; this matches the formula a_n = a + (n − 1)d = dn + (a − d).

These questions test understanding rather than computation: the sum grows quadratically, the term grows linearly, and each can be obtained from the other. Practise the subtraction carefully, especially the expansion of (n − 1)², which is the usual source of error.

Money that changes by a fixed amount each period is an AP, and the total paid or saved over a number of periods is the sum of the AP. These are the most common eight-mark applications.

Loan instalments. A sum of Rs 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year and show that the interests form an AP; find the interest at the end of 30 years. Interest each year = Rs 80, so the interest at the end of year 1, 2, 3, ... is 80, 160, 240, ..., an AP with a = 80, d = 80. After 30 years: 80 + 29 × 80 = 2400. Rs 2400.

A loan repaid in rising instalments. Jaspal Singh repays his total loan of Rs 118,000 by paying every month starting with the first instalment of Rs 1000, and increasing the instalment by Rs 100 every month. What amount will he pay in the 30th instalment, and what amount is still due after the 30th instalment? Instalments 1000, 1100, 1200, ...: a = 1000, d = 100. a₃₀ = 1000 + 29 × 100 = 3900. S₃₀ = 15(2000 + 2900) = 15 × 4900 = 73,500. Still due: 118,000 − 73,500 = Rs 44,500.

Paying with decreasing instalments. A man repays a loan of Rs 3250 by paying Rs 20 in the first month and increasing the payment by Rs 15 every month. How long will it take? S_n = (n/2)[40 + 15(n − 1)] = 3250, n(15n + 25) = 6500, 15n² + 25n − 6500 = 0, 3n² + 5n − 1300 = 0, (3n + 65)(n − 20) = 0, n = 20 months.

Savings. A person saves Rs 32 in the first month, Rs 36 in the second, Rs 40 in the third, and so on. In how many months will he save Rs 2000? (n/2)[64 + 4(n − 1)] = 2000, n(4n + 60) = 4000, n² + 15n − 1000 = 0, (n + 40)(n − 25) = 0, n = 25 months.

Annual salary with increments. An employee earns Rs 3,00,000 in the first year and gets an increment of Rs 15,000 every year. What is the total earned in 10 years? a = 300000, d = 15000, n = 10: S = 5(600000 + 135000) = 5 × 735000 = Rs 36,75,000.

Prizes. A sum of Rs 700 is to be used to give seven cash prizes to students, each prize Rs 20 less than the one before it. Find each prize. Seven terms in AP with sum 700 and d = −20: (7/2)(2a − 120) = 700, 2a − 120 = 200, a = 160. Prizes: 160, 140, 120, 100, 80, 60, 40.

Penalty for delay. A contract stipulates a penalty of Rs 200 for the first day of delay, Rs 250 for the second, Rs 300 for the third, and so on. How much must a contractor pay if he delays the work by 30 days? a = 200, d = 50, n = 30: S = 15(400 + 29 × 50) = 15(400 + 1450) = 15 × 1850 = Rs 27,750.

Note on simple versus compound interest. Simple interest produces an AP (the same interest every year); compound interest does not, because each year's interest is larger than the last by a growing amount. Questions sometimes ask which of several money situations is an AP; the test is whether the change each period is the same fixed amount.

In each problem, name the AP explicitly (a = ..., d = ...), decide whether the question asks for a term (a_n) or a total (S_n), apply the right formula, and answer in rupees.

Objects arranged in rows that grow or shrink by a fixed number form an AP, and the total number of objects is the sum.

Logs. 200 logs are stacked with 20 in the bottom row, 19 in the next, 18 in the next, and so on. In how many rows are the logs placed, and how many are in the top row? a = 20, d = −1, S_n = 200: (n/2)[40 − (n − 1)] = 200, n(41 − n) = 400, n² − 41n + 400 = 0, (n − 16)(n − 25) = 0, n = 16 or 25. If n = 25, the 25th row would have 20 − 24 = −4 logs, impossible; so n = 16, and the top row has 20 − 15 = 5 logs. This problem shows that an algebraically valid root can be physically impossible and must be rejected with a reason.

Seats in a hall. A hall has 30 seats in the first row, 32 in the second, 34 in the third, and so on for 25 rows. How many seats are there? a = 30, d = 2, n = 25: S = (25/2)(60 + 48) = 25 × 54 = 1350 seats.

Trees planted. In a school, students of each section of each class plant as many trees as their class number, from Class 1 to Class 12, with three sections in each class. How many trees are planted? Each class plants 3 × (class number): 3, 6, 9, ..., 36; S = (12/2)(3 + 36) = 6 × 39 = 234 trees.

Ladder rungs. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2½ m apart, what length of wood is needed for the rungs? Number of rungs = 250/25 + 1 = 11. Lengths form an AP from 45 to 25 with 11 terms: S = (11/2)(45 + 25) = 11 × 35 = 385 cm.

Potato race. A bucket is placed at the start and 10 potatoes are placed 3 m apart in a line, the first 5 m from the bucket. A competitor picks up each potato in turn and runs back to drop it in the bucket. How far does she run in all? For the potato at distance x she runs 2x. Distances: 2 × 5, 2 × 8, 2 × 11, ... = 10, 16, 22, ...; a = 10, d = 6, n = 10: S = 5(20 + 54) = 370 m.

Semicircles in a spiral. A spiral is made of thirteen consecutive semicircles with centres alternately at A and B, starting with radius 0.5 cm and increasing by 0.5 cm each time. Its total length (taking π = 22/7): the lengths are πr for r = 0.5, 1.0, 1.5, ..., 6.5, so the total is π(0.5 + 1.0 + ... + 6.5) = π × (13/2)(0.5 + 6.5) = π × 45.5 = (22/7)(45.5) = 143 cm.

Houses numbered on a street. Houses on a row are numbered 1 to 49. Show there is a house number x such that the sum of the numbers before it equals the sum of the numbers after it. Sum before x is (x − 1)x/2; sum after is 49 × 50/2 − x(x + 1)/2 = 1225 − x(x + 1)/2. Equate: x(x − 1) = 2450 − x(x + 1), so 2x² = 2450, x² = 1225, x = 35. House number 35.

Triangular patterns. A triangular pattern of dots with 1, 2, 3, ... dots per row for 20 rows has 20 × 21/2 = 210 dots. And the number of handshakes among n people, n(n − 1)/2, is the sum 1 + 2 + ... + (n − 1).

The key to each problem is to write down the first few terms explicitly, identify a and d, decide whether the number of rows or the total is asked, and reject roots that give a negative or fractional count.

Some examination questions ask you to prove a general statement about an AP rather than to compute a number. The tools are the same two formulas, used with letters instead of numbers.

Result 1: if the pth term is q and the qth term is p, then the (p + q)th term is 0. a + (p − 1)d = q and a + (q − 1)d = p. Subtract: (p − q)d = q − p, so d = −1 (for p ≠ q). Then a = q − (p − 1)(−1) = p + q − 1. The (p + q)th term is a + (p + q − 1)d = (p + q − 1) − (p + q − 1) = 0.

Result 2: if m times the mth term equals n times the nth term, the (m + n)th term is 0. m[a + (m − 1)d] = n[a + (n − 1)d]. Expand: ma + m(m − 1)d = na + n(n − 1)d, so (m − n)a + [m² − m − n² + n]d = 0, so (m − n)a + (m − n)(m + n − 1)d = 0. Divide by m − n ≠ 0: a + (m + n − 1)d = 0, which is the (m + n)th term.

Result 3: if S_p = S_q with p ≠ q, then S_p+q = 0. Proved in an earlier section by the same factoring of p − q.

Result 4: the sum of the first n odd natural numbers is n². 1 + 3 + ... + (2n − 1) = (n/2)(1 + 2n − 1) = n². And the sum of the first n even natural numbers is 2 + 4 + ... + 2n = (n/2)(2 + 2n) = n(n + 1).

Result 5: the ratio of the sums of two APs. If the sums of the first n terms of two APs are in the ratio (7n + 1) : (4n + 27), find the ratio of their mth terms. For an AP, S_n/n = a + (n − 1)d/2, which equals the term a_k with (k − 1) = (n − 1)/2, that is k = (n + 1)/2. So the ratio of the mth terms is the ratio of the sums with n = 2m − 1: (7(2m − 1) + 1)/(4(2m − 1) + 27) = (14m − 6)/(8m + 23). Check with m = 1: the first terms are in the ratio 8/31, and S₁ ratio is (7 + 1)/(4 + 27) = 8/31. The trick: replace n by 2m − 1.

Result 6: if a, b, c are in AP, then so are b + c, c + a, a + b. Since 2b = a + c, check whether 2(c + a) = (b + c) + (a + b): the right side is a + c + 2b = a + c + a + c = 2(a + c). Yes. Similarly, if a, b, c are in AP then a², b², c² need not be, but 1/(bc), 1/(ca), 1/(ab) are, because dividing an AP by the non-zero constant abc keeps it an AP.

Result 7: inserting arithmetic means. To insert k numbers between a and b so that the whole list is an AP, we need k + 2 terms with first term a and last term b, so d = (b − a)/(k + 1). Insert 3 numbers between 8 and 26: d = 18/4 = 4.5; the numbers are 12.5, 17, 21.5. Insert 5 numbers between 1 and 19: d = 3; the numbers are 4, 7, 10, 13, 16.

Result 8: the middle term. The sum of a finite AP with an odd number of terms is the number of terms times the middle term, because the middle term is the average of the first and last: S_2k+1 = (2k + 1)a_k+1. For 3, 7, 11, 15, 19: five terms, middle 11, sum 55. If the sum of three consecutive terms of an AP is 51, the middle one is 17.

In a proof, write the given in symbols, state what is to be shown, manipulate one side or both towards the target, and finish with the conclusion. The factor (p − q) or (m − n) appears in almost every such proof and is the thing to look for.

A group of problems combines the AP formulas with number facts.

Sums of multiples. Find the sum of all natural numbers between 1 and 100 that are divisible by 6: 6, 12, ..., 96, n = 16, S = 8(6 + 96) = 816. Find the sum of all three-digit numbers divisible by 13: 104, 117, ..., 988; n = (988 − 104)/13 + 1 = 69; S = (69/2)(104 + 988) = 69 × 546 = 37674. Find the sum of numbers between 1 and 100 divisible by 2 or 5: divisible by 2 is 2 + ... + 100 = 2550; by 5 is 5 + ... + 100 = 1050; by 10 (counted twice) is 10 + ... + 100 = 550; total 2550 + 1050 − 550 = 3050. Find the sum of integers from 1 to 100 not divisible by 3 or 5: total 5050 minus multiples of 3 (1683) minus multiples of 5 (1050) plus multiples of 15 (315) gives 2632.

Remainders. Find the sum of all two-digit numbers which when divided by 4 leave remainder 1: 13, 17, ..., 97; n = (97 − 13)/4 + 1 = 22; S = 11(13 + 97) = 1210. Numbers between 100 and 200 that leave remainder 3 on division by 7: 101, 108, ..., 199; n = 15; S = (15/2)(101 + 199) = 2250.

Odd or even terms. The sum of the odd-positioned terms of an AP: the 1st, 3rd, 5th, ... terms themselves form an AP with common difference 2d. In 3, 8, 13, ..., 253 (51 terms), the sum of the terms in odd positions is 3 + 13 + 23 + ... + 253 (26 terms, d = 10): 13(3 + 253) = 3328. Sum of the terms in even positions: 8 + 18 + ... + 248 (25 terms): (25/2)(8 + 248) = 3200. Check: 3328 + 3200 = 6528 = (51/2)(3 + 253) = 6528.

Mixed conditions on terms. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th is 44. Find the first three terms. a + 3d + a + 7d = 24 gives a + 5d = 12; a + 5d + a + 9d = 44 gives a + 7d = 22; subtracting, 2d = 10, d = 5, a = −13. The AP is −13, −8, −3, ... The 4th term of an AP is 11 and the sum of the 5th and 7th terms is 34: a + 3d = 11, 2a + 10d = 34 so a + 5d = 17; d = 3, a = 2; the AP is 2, 5, 8, ...; its 20th term is 59. The sum of the first four terms is 40 and the sum of the first 14 terms is 280; find the sum of the first n terms: 2(2a + 3d) = 40 gives 2a + 3d = 20; 7(2a + 13d) = 280 gives 2a + 13d = 40; 10d = 20, d = 2, a = 7; S_n = (n/2)(14 + 2(n − 1)) = n(n + 6) = n² + 6n.

Terms and sums together. The first term of an AP is 5, the last is 45 and the sum is 400: n = 16, d = 8/3 (found earlier). The first and last terms are 17 and 350 and d = 9: n = (350 − 17)/9 + 1 = 38, S = 19(17 + 350) = 6973. The sum of the first 14 terms is 1050 and the first term is 10: 7(20 + 13d) = 1050, 20 + 13d = 150, d = 10, a₂₀ = 10 + 190 = 200.

Reverse problems. If the 3rd and 9th terms are 4 and −8, which term is 0? Found earlier: the 5th. If the 10th term is 52 and the 17th is 20 more than the 13th, find the AP: 4d = 20, d = 5, a + 45 = 52, a = 7; AP 7, 12, 17, ... The 24th term is twice the 10th; show the 72nd is twice the 34th: a + 23d = 2(a + 9d) gives a = 5d; then a₇₂ = 5d + 71d = 76d and a₃₄ = 5d + 33d = 38d, so a₇₂ = 2a₃₄.

These problems reward a systematic habit: translate each condition into an equation in a and d, solve the pair, and then compute what is asked. Two conditions on terms always give two linear equations, exactly as in the chapter on pairs of linear equations.

One-mark questions. Give the common difference of a given AP. Write the next term. Find the nth term formula of a simple AP (e.g. 3, 7, 11: a_n = 4n − 1). Find a specified term. State whether a list is an AP. Find the sum of the first n natural numbers or first n odd numbers. Find the value of x for which 2x, x + 10, 3x + 2 are in AP (2(x + 10) = 5x + 2, x = 6). State the 10th term from the end of an AP with given l and d.

Two-mark questions. Find a and d from two given terms. Which term of an AP is a given number, or show that a given number is not a term. Find the number of terms of a finite AP. Find the sum of a short finite AP. Insert arithmetic means.

Four-mark questions. Find how many terms give a given sum (quadratic in n, with both roots discussed). Find the AP given the sum of some terms and a product or another sum. Find the nth term from a sum formula. Sum of multiples in a range. A money problem asking for a term and a total.

Eight-mark questions. The logs, the potato race, the ladder, the spiral, the houses numbered 1 to 49, or a proof such as a_m+n = 0 or S_p+q = 0. Full formulation, solution, rejection of impossible roots and a stated answer are all marked.

Common errors. (1) Using n instead of n − 1 in the nth term formula; a_n = a + (n − 1)d, so the 10th term has nine d's. (2) Counting the number of terms as (l − a)/d instead of (l − a)/d + 1. (3) Wrong sign of d for a decreasing AP. (4) In the sum formula, forgetting the n/2 or writing (n/2)(a + l) with a_n that was computed wrongly. (5) Accepting a fractional or negative n as an answer, or rejecting a second valid integer n without checking whether it is genuinely valid (it is valid when the intervening terms sum to zero, and invalid when it produces a negative count of objects). (6) In a_n = S_n − S_n−1, expanding (n − 1)² wrongly. (7) In money problems, confusing the amount in the nth period (a term) with the total up to the nth period (a sum). (8) Treating compound interest as an AP. (9) Taking three numbers in AP as a, a + d, a + 2d when a − d, a, a + d makes the sum condition trivial. (10) Arithmetic slips in large multiplications; keep the n/2 factor until the last step to reduce the size of numbers.

Summary. An AP is a, a + d, a + 2d, ...; d is any term minus the previous. a_n = a + (n − 1)d. S_n = (n/2)[2a + (n − 1)d] = (n/2)(a + l). a_n = S_n − S_n−1. Three terms: a − d, a, a + d. Middle term is the mean of neighbours. Everything else is translation and algebra.