Chapter 10 — Progressions

Look at these lists of numbers. 5, 10, 15, 20, 25, … : each number is 5 more than the one before it. 1, 2, 4, 8, 16, … : each number is twice the one before. 1, 4, 9, 16, 25, … : the squares of 1, 2, 3, 4, 5. 2, 3, 5, 7, 11, … : the primes. Each is a sequence, an ordered list of numbers in which every position has a definite number. The number at position 1 is the first term, at position 2 the second term, and in general the number at position n is the nth term, written a_n. A sequence that follows a rule allowing every term to be found is called a progression.

The point of the chapter is this: if the rule is known, we never need to write the whole list. We can find the 100th term of 5, 10, 15, … directly, and the sum of the first 100 terms too, without adding a hundred numbers. Two rules are so common that they have names and formulas.

In an arithmetic progression (AP) each term is obtained from the previous one by adding a fixed number. In 5, 10, 15, 20 the fixed number is 5; in 100, 90, 80, 70 it is −10; in 3, 3, 3, 3 it is 0. In a geometric progression (GP) each term is obtained by multiplying the previous one by a fixed number. In 1, 2, 4, 8 it is 2; in 81, 27, 9, 3 it is 1/3; in 2, −6, 18, −54 it is −3.

Everyday situations produce these. A taxi charges ₹20 for the first kilometre and ₹8 for every kilometre after that: the fares for 1, 2, 3, 4 km are 20, 28, 36, 44, an AP with difference 8. A sum of ₹1000 at 10% compound interest becomes 1100, 1210, 1331, … an AP would be 1100, 1200, 1300 under simple interest, but compound interest gives a GP with ratio 1.1. The number of bacteria doubling every hour, the length of a spring under successive equal loads, the seats in successive rows of a stadium, the depth of a well dug at a fixed cost per metre with a fixed increase per metre — all are progressions.

Not every sequence is a progression of these kinds. The squares 1, 4, 9, 16 have differences 3, 5, 7 that are not constant, and ratios 4, 9/4, 16/9 that are not constant either; so they are neither an AP nor a GP, though the differences themselves form an AP. The first skill is to test a given list correctly: compute the differences of consecutive terms, and if they are all equal it is an AP; compute the ratios, and if they are all equal it is a GP.

An arithmetic progression is a sequence in which each term, except the first, is obtained by adding a fixed number to the preceding term. The fixed number is the common difference, written d, and the first term is written a. Then the sequence is

a, a + d, a + 2d, a + 3d, …

This is the general form of an AP. The common difference can be positive (an increasing AP, such as 2, 5, 8, 11 with d = 3), negative (a decreasing AP, such as 20, 15, 10, 5 with d = −5), or zero (a constant AP, such as 7, 7, 7, 7).

To find d from a given AP, subtract any term from the next one: d = a₂ − a₁ = a₃ − a₂, and so on. It must be the next minus the previous, not the other way round; for 20, 15, 10, d = 15 − 20 = −5. To check whether a list is an AP, compute all consecutive differences and verify they are equal. The list 1, 3, 9, 27 has differences 2, 6, 18 and is not an AP. The list √2, √8, √18, √32 is √2, 2√2, 3√2, 4√2 with d = √2, so it is an AP even though it does not look like one at first.

An AP may be finite, with a definite last term, such as the multiples of 3 between 10 and 40: 12, 15, …, 39; or infinite, going on for ever, such as the odd numbers 1, 3, 5, 7, …. A finite AP has a last term, usually denoted l.

Writing an AP from a and d is direct: a = 10, d = 10 gives 10, 20, 30, 40; a = −2, d = 0 gives −2, −2, −2; a = 4, d = −3 gives 4, 1, −2, −5; a = 1/2, d = 1/2 gives 1/2, 1, 3/2, 2. Conversely, given the first few terms, a is the first and d is the difference.

Word situations: the cost of digging a well is ₹150 for the first metre and rises by ₹50 for each subsequent metre, so the costs of successive metres are 150, 200, 250, 300, … with a = 150, d = 50. The amount in an account with ₹10000 at 8% simple interest after each year is 10800, 11600, 12400, … with a = 10800, d = 800. The number of dots in a triangular pattern 1, 3, 6, 10 is not an AP (differences 2, 3, 4), a distinction the examination likes to test.

A useful trick for three or four unknown terms in an AP: take three terms as a − d, a, a + d and four terms as a − 3d, a − d, a + d, a + 3d. The sums then simplify to 3a and 4a, and the unknowns are found from one more condition.

In the general form a, a + d, a + 2d, a + 3d, … the first term has no d, the second has one d, the third has two d, the fourth has three d. The pattern is clear: the term at position n has (n − 1) copies of d added to a. Therefore

a_n = a + (n − 1)d

This is the formula for the nth term, also called the general term, of an AP. It has four quantities, a, d, n and a_n, and every problem gives three of them and asks for the fourth.

Finding a term. Find the 10th term of 2, 7, 12, …. Here a = 2, d = 5, n = 10, so a₁₀ = 2 + 9 × 5 = 47. Find the 20th term of 100, 90, 80, …: a = 100, d = −10, a₂₀ = 100 + 19(−10) = −90.

Finding the position of a term. Which term of 21, 18, 15, … is −81? Here a = 21, d = −3, a_n = −81. So 21 + (n − 1)(−3) = −81, so −3(n − 1) = −102, n − 1 = 34, n = 35. The 35th term. If solving gives n that is not a positive integer, the number is not a term: is 0 a term of 21, 18, 15, …? 21 − 3(n − 1) = 0 gives n − 1 = 7, n = 8, so yes, 0 is the 8th term. Is 301 a term of 5, 11, 17, …? 5 + 6(n − 1) = 301 gives n − 1 = 296/6, not an integer, so no.

Counting terms. How many terms are in 7, 13, 19, …, 205? a = 7, d = 6, l = 205: 7 + 6(n − 1) = 205, n − 1 = 33, n = 34 terms. How many two-digit numbers are divisible by 3? They are 12, 15, …, 99: 12 + 3(n − 1) = 99, n − 1 = 29, n = 30.

Finding a and d from two terms. If the 3rd term is 5 and the 7th is 9, then a + 2d = 5 and a + 6d = 9. Subtracting, 4d = 4, d = 1, a = 3. The AP is 3, 4, 5, …. This pair-of-equations method appears in almost every examination.

Terms from the end. The nth term from the end of a finite AP with last term l is l − (n − 1)d. The 11th term from the last of 10, 7, 4, …, −62 is −62 − 10(−3) = −32. Equivalently, reverse the AP (first term −62, difference 3) and use the usual formula.

Note that a_n = a + (n − 1)d is a linear expression in n, so a sequence is an AP exactly when its nth term is of the form pn + q; then d = p and a = p + q. The sequence a_n = 3n + 2 is an AP with a = 5, d = 3; the sequence a_n = n² + 1 is not.

The nth-term formula turns many word problems into one-line computations once the AP is identified. The discipline is: write the first few terms from the story, identify a and d, decide which quantity is asked, and apply a_n = a + (n − 1)d.

Salary with increments. Subba Rao started work in 1995 at a monthly salary of ₹5000 and received an increment of ₹200 each year. In which year did his salary reach ₹7000? The salaries in successive years form the AP 5000, 5200, 5400, … with a = 5000, d = 200. We need 5000 + 200(n − 1) = 7000, so n − 1 = 10, n = 11. The 11th year of service is 1995 + 10 = 2005.

Savings. Ramkali saved ₹5 in the first week and increased her weekly saving by ₹1.75 each week. In which week will her saving be ₹20.75? a = 5, d = 1.75, a_n = 20.75: 5 + 1.75(n − 1) = 20.75, 1.75(n − 1) = 15.75, n − 1 = 9, n = 10. The 10th week.

Multiples in a range. How many multiples of 4 lie between 10 and 250? The first is 12, the last is 248, d = 4: 12 + 4(n − 1) = 248, n − 1 = 59, n = 60.

Common terms. For what value of n are the nth terms of 63, 65, 67, … and 3, 10, 17, … equal? 63 + 2(n − 1) = 3 + 7(n − 1), so 60 = 5(n − 1), n − 1 = 12, n = 13. Both have 13th term 87.

Three-digit numbers divisible by 7. The first is 105, the last is 994, d = 7: 105 + 7(n − 1) = 994, n − 1 = 127, n = 128.

A term as a multiple of another. If the 17th term exceeds the 10th by 7, find d: (a + 16d) − (a + 9d) = 7 gives 7d = 7, d = 1. If the 11th term is three times the 3rd, then a + 10d = 3(a + 2d), so 4d = 2a, a = 2d; the AP is 2d, 3d, 4d, ….

Terms equidistant from the ends. In a finite AP, the sum of the kth term from the beginning and the kth term from the end is always a + l. For 3, 7, 11, …, 99: the 5th from the start is 19, the 5th from the end is 83, and 19 + 83 = 102 = 3 + 99. This is the fact behind the sum formula of the next topic.

Checking the answer against the story matters here too: n must be a positive integer, and a year, a week or a count must make sense. If the arithmetic gives n = 9.5, either the number is not a term, or an error has crept in — re-read the problem.

A story is told of the young Gauss, asked to add the numbers 1 to 100, who wrote the list forwards and backwards, noticed that every pair added to 101, and answered 100 × 101 ÷ 2 = 5050 at once. The same idea gives the sum of any AP.

Let S_n be the sum of the first n terms of the AP a, a + d, a + 2d, …, with nth term a + (n − 1)d. Write

S_n = a + (a + d) + (a + 2d) + … + [a + (n − 1)d]

and the same sum reversed,

S_n = [a + (n − 1)d] + [a + (n − 2)d] + … + (a + d) + a.

Add the two lines term by term. Each vertical pair adds to a + [a + (n − 1)d] = 2a + (n − 1)d, and there are n such pairs. So 2S_n = n[2a + (n − 1)d], and

S_n = n/2 [2a + (n − 1)d]

If the last term l = a + (n − 1)d is known, then 2a + (n − 1)d = a + l, and the formula takes the shorter form

S_n = n/2 (a + l)

which reads: the sum is the number of terms times the average of the first and last terms.

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, l = 10, d = −2; number of terms from 34 − 2(n − 1) = 10, n = 13; S = 13/2 (34 + 10) = 13 × 22 = 286. Sum of the first 1000 positive integers: n/2 (1 + 1000) = 500500.

Two special sums are worth remembering as instances of the formula: the sum of the first n natural numbers is n(n + 1)/2, and the sum of the first n odd numbers 1 + 3 + 5 + … + (2n − 1) is n² (a = 1, d = 2: n/2[2 + 2(n − 1)] = n/2 × 2n = n²).

The formula also gives the nth term back: since S_n − S_n−1 is the nth term added, a_n = S_n − S_n−1. If S_n = 4n − n², then a₁ = S₁ = 3, a₂ = S₂ − S₁ = 4 − 3 = 1, and a_n = (4n − n²) − [4(n − 1) − (n − 1)²] = 5 − 2n; the AP is 3, 1, −1, …, and the 10th term is −15.

The sum formula contains four quantities S_n, n, a and d, and the harder problems give the sum and ask for one of the others. Since n appears squared in the formula, these problems often lead to a quadratic in n, whose positive integer root is the answer.

Finding n from the sum. How many terms of 24, 21, 18, … must be taken so that the 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. Both n = 4 and n = 13 are valid: the first four terms 24 + 21 + 18 + 15 = 78, and the terms from the 5th to the 13th (12, 9, 6, 3, 0, −3, −6, −9, −12) add to zero, so the first 13 terms also add to 78. When two positive integer answers appear, give both and explain.

Finding d from two sums. If the sum of the first 7 terms 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. Then S_n = n/2[2 + 2(n − 1)] = n². Indeed 49 = 7² and 289 = 17².

Finding a from a sum and a term. The first term is 5, the last term is 45 and the sum is 400. Find n and d. S = n/2(a + l): 400 = n/2 × 50, n = 16. Then 45 = 5 + 15d, d = 8/3.

Sum of a specified range of terms. The sum of the 11th to the 20th terms is S₂₀ − S₁₀. For 3, 8, 13, …: S₂₀ = 10[6 + 95] = 1010, S₁₀ = 5[6 + 45] = 255, so the required sum is 755.

A condition on sums of terms. If the sum of the first n terms of an AP is 3n² + 5n and its mth term is 164, find m. a_m = S_m − S_m−1 = 3m² + 5m − [3(m − 1)² + 5(m − 1)] = 6m + 2. So 6m + 2 = 164, m = 27.

Sums of multiples. The sum of the first 40 positive integers divisible by 6: 6 + 12 + … + 240 = 6(1 + 2 + … + 40) = 6 × 820 = 4920. The sum of all three-digit multiples of 7: 105 + … + 994, 128 terms, S = 64 × (105 + 994) = 64 × 1099 = 70336.

When the equation in n is quadratic, always reject a negative or fractional root, and if both roots are positive integers, check the terms between them: they must add to zero for both to be valid, which happens only when the AP passes through zero symmetrically.

The sum formula answers questions about totals accumulated over time or over a stack, whenever the amounts form an AP. The examination sets these as three- or four-mark problems, and the marks are for identifying a, d and n correctly and stating the formula before computing.

A contractor's penalty. A contract specifies a penalty for delay: ₹200 for the first day, ₹250 for the second, ₹300 for the third, and so on, each day ₹50 more than the previous. How much must the contractor pay if the work is delayed by 30 days? The daily penalties form an AP with a = 200, d = 50, n = 30. S₃₀ = 15[400 + 29 × 50] = 15[400 + 1450] = 15 × 1850 = ₹27750.

Prizes decreasing by a fixed amount. A sum of ₹700 is to be given as seven cash prizes, each ₹20 less than the preceding one. Find the prizes. Here S₇ = 700, d = −20, n = 7: 7/2[2a − 120] = 700, 2a − 120 = 200, a = 160. The prizes are 160, 140, 120, 100, 80, 60, 40.

Trees planted by sections. A school has three sections in each of classes 1 to 12, and each section of class k plants k trees. Each class plants 3k trees, so the total is 3(1 + 2 + … + 12) = 3 × 78 = 234 trees.

Logs in a stack. 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. 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.

Savings that grow each year. A person saves ₹3200 in the first year and increases the saving by ₹400 each year. In how many years will the total savings reach ₹80000? n/2[6400 + 400(n − 1)] = 80000, n(6000 + 400n) = 160000, 400n² + 6000n − 160000 = 0, n² + 15n − 400 = 0, (n + 25)(n − 16) = 0, n = 16 years.

Potatoes in a race. Ten potatoes are placed 3 m apart in a line, the first 5 m from a bucket. A competitor runs from the bucket to each potato and back with it, one at a time. The distances run are 2 × 5, 2 × 8, 2 × 11, …: an AP with a = 10, d = 6, n = 10. Total = 5[20 + 54] = 370 m.

A spiral of semicircles. Semicircles of radii 0.5, 1.0, 1.5, … cm are drawn one after another, thirteen of them. The lengths are π × 0.5, π × 1.0, …, so the total length is π(0.5 + 1.0 + … + 6.5) = π × 13/2 × (0.5 + 6.5) = π × 45.5 = 143 cm with π = 22/7.

If three numbers a, b, c are in AP, the middle one is the average of the other two, because b − a = c − b gives b = (a + c)/2. This middle term is the arithmetic mean (AM) of a and c. The AM of 3 and 11 is 7, and 3, 7, 11 is an AP. Conversely, a, b, c are in AP if and only if 2b = a + c, a test used often in proofs.

More generally, we can insert any number of terms between two given numbers so that the whole becomes an AP. To insert k terms between p and q, the new AP has k + 2 terms, first term p and last term q, so q = p + (k + 1)d, giving d = (q − p)/(k + 1).

Example. Insert 5 terms between 4 and 22. Then d = (22 − 4)/6 = 3, and the terms are 7, 10, 13, 16, 19. Insert 3 terms between −2 and 10: d = 12/4 = 3, giving 1, 4, 7.

Finding an unknown from the AP condition. For what value of x are 2x, x + 10 and 3x + 2 in AP? The middle term is the mean: 2(x + 10) = 2x + 3x + 2, so 2x + 20 = 5x + 2, 3x = 18, x = 6. The terms are 12, 16, 20. If k + 9, 2k − 1 and 2k + 7 are in AP, 2(2k − 1) = (k + 9) + (2k + 7), 4k − 2 = 3k + 16, k = 18.

Middle term of a finite AP. If a finite AP has an odd number of terms, its middle term is the average of the first and last terms, and the sum equals the middle term times the number of terms. The AP 6, 10, 14, …, 102 has n given by 6 + 4(n − 1) = 102, n = 25; the middle (13th) term is 6 + 12 × 4 = 54 = (6 + 102)/2, and the sum is 25 × 54 = 1350. If the number of terms is even, there are two middle terms and their average is (a + l)/2.

Terms symmetric about the middle. Choosing three unknown terms as a − d, a, a + d exploits the AM: a is the mean, so the sum 3a fixes a immediately. The angles of a triangle in AP with the greatest twice the least: a − d + a + a + d = 180 gives a = 60; a + d = 2(a − d) gives 3d = a = 60, d = 20; the angles are 40°, 60°, 80°. Four terms with sum 20 and sum of squares 120: a − 3d, a − d, a + d, a + 3d give 4a = 20, a = 5, and 4a² + 20d² = 120 gives 100 + 20d² = 120, d = ±1; the numbers are 2, 4, 6, 8.

The AM also explains why the sum formula is n(a + l)/2: the average of all the terms of an AP equals the average of the first and last, and a sum is the number of terms times their average.

A geometric progression is a sequence in which each term, except the first, is obtained by multiplying the preceding term by a fixed nonzero number. The fixed number is the common ratio, written r, and with first term a the sequence is

a, ar, ar², ar³, …

This is the general form of a GP. Neither a nor r can be zero: if a = 0 every term is 0 and there is no ratio, and if r = 0 the sequence collapses after the first term.

To find r, divide any term by the previous one: r = a₂/a₁ = a₃/a₂. For 2, 6, 18, 54, r = 3. For 64, 32, 16, 8, r = 1/2. For 5, −10, 20, −40, r = −2, and the terms alternate in sign. For 3, 3, 3, 3, r = 1, a constant GP (which is also an AP with d = 0). For 1/3, 1/9, 1/27, r = 1/3.

To test whether a list is a GP, compute all consecutive ratios and check they are equal. 4, 8, 12, 16 has ratios 2, 3/2, 4/3 — not a GP (it is an AP). 1, −1, 1, −1 has ratio −1 throughout — a GP. √2, 2, 2√2, 4 has ratio √2 — a GP. −3, −6, −12 has ratio 2 — a GP whose terms are all negative.

Situations producing GPs: compound interest, where each year's amount is the previous times (1 + R/100); a population growing by a fixed percentage; the halving of a quantity such as a radioactive sample or the height of a bouncing ball, each bounce reaching a fixed fraction of the previous height; the number of ancestors in each generation (2, 4, 8, 16); the successive folds of a paper doubling its thickness; the lengths of the sides of squares built inside squares by joining midpoints (ratio 1/√2). A sum of ₹500 at 10% compound interest gives 500, 550, 605, 665.50, … with r = 1.1.

Writing a GP from a and r: a = 4, r = 3 gives 4, 12, 36, 108; a = √5, r = 1/5 gives √5, √5/5, √5/25; a = 81, r = −1/3 gives 81, −27, 9, −3, 1.

A GP grows or decays much faster than an AP. Compare 1, 3, 5, 7, … (AP, d = 2) with 1, 3, 9, 27, … (GP, r = 3): by the 10th term the AP has reached 19 and the GP 19683. When |r| > 1 the terms grow without bound; when |r| < 1 they shrink towards zero; when r is negative they alternate in sign. This qualitative behaviour is the first thing to note about any GP.

Three numbers in GP are conveniently taken as a/r, a, ar, whose product is a³; this parallels the a − d, a, a + d trick for APs.

In the general form a, ar, ar², ar³, … the first term has r⁰, the second r¹, the third r², so the nth term has r to the power n − 1:

a_n = a · rⁿ⁻¹

Finding a term. The 10th term of 5, 25, 125, …: a = 5, r = 5, a₁₀ = 5 × 5⁹ = 5¹⁰. The 7th term of 2, −6, 18, …: r = −3, a₇ = 2 × (−3)⁶ = 2 × 729 = 1458. The 6th term of 64, 32, 16, …: r = 1/2, a₆ = 64 × (1/2)⁵ = 64/32 = 2.

Finding the position of a term. Which term of 2, 2√2, 4, … is 128? a = 2, r = √2, so 2(√2)ⁿ⁻¹ = 128, (√2)ⁿ⁻¹ = 64 = 2⁶ = (√2)¹², so n − 1 = 12, n = 13. Which term of 5, 10, 20, … is 1280? 5 × 2ⁿ⁻¹ = 1280, 2ⁿ⁻¹ = 256 = 2⁸, n = 9.

Finding a and r from two terms. If the 4th term is 24 and the 7th is 192: ar³ = 24 and ar⁶ = 192. Dividing, r³ = 8, r = 2, so a = 24/8 = 3. The GP is 3, 6, 12, 24, …. Dividing one equation by the other is the standard method, because it eliminates a at once.

An exponent to solve. If the 3rd term of a GP is 4, the product of the first five terms is a · ar · ar² · ar³ · ar⁴ = a⁵r¹⁰ = (ar²)⁵ = 4⁵ = 1024, without knowing a or r separately. This uses the fact that the middle term of five terms in GP is the geometric mean of the whole set.

Word problems. A ball dropped from 16 m rebounds each time to half the previous height. The heights after successive bounces are 8, 4, 2, …: a = 8, r = 1/2, and the height after the 5th bounce is 8 × (1/2)⁴ = 0.5 m. A city of 8 lakh doubles its population every 25 years; after n such periods it has 8 × 2ⁿ lakh, so after 100 years (4 periods) it has 128 lakh. A sum at 8% compound interest: amount after n years = P(1.08)ⁿ, a GP in n with r = 1.08.

Equal terms of two GPs. For what n are the nth terms of 2, 8, 32, … and 1024, 512, 256, … equal? 2 × 4ⁿ⁻¹ = 1024 × (1/2)ⁿ⁻¹. Write both sides in powers of 2: 2 × 2²⁽ⁿ⁻¹⁾ = 2¹⁰ × 2⁻⁽ⁿ⁻¹⁾, so 1 + 2n − 2 = 10 − n + 1, 3n = 12, n = 4. Check: both 4th terms are 128.

The examination's Class 10 treatment of GPs stops at the nth term; the sum of a GP is Intermediate work. What is examined is the correct identification of r (including negative and fractional ratios), the nth-term formula, and the algebra of powers needed to solve for n.

Many examination questions present a situation and ask which progression models it, or present a list and ask for its type. The distinction is simple: in an AP the difference between consecutive terms is constant, so the sequence changes by a fixed amount; in a GP the ratio is constant, so the sequence changes by a fixed percentage or factor.

Situations that are APs: simple interest amounts year by year; a salary with a fixed annual increment; a taxi meter; the cost of digging each successive metre of a well when the rate rises by a fixed sum; the seats in rows of a hall increasing by a fixed number; the perimeters of squares whose sides increase by a fixed length; a fixed penalty increase per day; the amount of air remaining in a cylinder when a fixed volume is removed each stroke.

Situations that are GPs: compound interest; population growth at a fixed percentage; the depreciation of a machine losing a fixed percentage of value each year (r = 1 − rate); bacteria doubling; the amount of air remaining when a pump removes a fixed fraction each stroke (r = 3/4 if a quarter is removed); the areas of squares built by joining midpoints (r = 1/2); the heights of a bouncing ball; the number of persons receiving a chain message when each sends it to a fixed number of others.

The air-pump example is a good test of understanding: if each stroke removes 1/4 of the air remaining, the amounts left are V, 3V/4, 9V/16, … a GP; if each stroke removes a fixed 1/4 of the original volume, the amounts are V, 3V/4, V/2, V/4, 0 — an AP that ends.

Sequences that are neither: squares, cubes, triangular numbers, primes, Fibonacci numbers. The differences of the squares 1, 4, 9, 16 form the AP 3, 5, 7, and the ratios of consecutive Fibonacci numbers approach a constant but never equal it. A question asking "is this an AP?" is answered by computing at least two differences, and the answer "no, because the differences 3 and 5 are unequal" earns the mark.

A single sequence can be both: a constant sequence 7, 7, 7 is an AP with d = 0 and a GP with r = 1. No other sequence is both, because a linear function of n and an exponential function of n agree at most at two points.

Beyond computation, the examination asks short proofs about progressions. These use only the nth-term and sum formulas and a little algebra, and the key is to write general terms with letters and manipulate them exactly.

If the pth term of an AP is q and the qth term is p, show that the (p + q)th term is 0. Let the AP have first term a and difference d. Then a + (p − 1)d = q and a + (q − 1)d = p. Subtracting, (p − q)d = q − p, so d = −1 (assuming 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.

If m times the mth term equals n times the nth term, show that the (m + n)th term is 0. m[a + (m − 1)d] = n[a + (n − 1)d]. So (m − n)a + [m(m − 1) − n(n − 1)]d = 0. Since m² − m − n² + n = (m − n)(m + n) − (m − n) = (m − n)(m + n − 1), we get (m − n)[a + (m + n − 1)d] = 0. As m ≠ n, a + (m + n − 1)d = 0, which is the (m + n)th term.

Sum of terms equidistant from the ends. In a finite AP with n terms, a_k + a_n−k+1 = [a + (k − 1)d] + [a + (n − k)d] = 2a + (n − 1)d = a + l, independent of k.

If a, b, c are in AP, then b + c, c + a, a + b are in AP. Given 2b = a + c. The proposed middle term is c + a, and (b + c) + (a + b) = a + c + 2b = 2(c + a) using 2b = a + c. So twice the middle equals the sum of the outer two, and the three are in AP. Similarly a², b², c² in AP implies 1/(b + c), 1/(c + a), 1/(a + b) in AP: the condition 2/(c + a) = 1/(b + c) + 1/(a + b) simplifies, after multiplying out, to 2b² = a² + c².

If the sums of the first p, q and r terms are P, Q, R, show that (P/p)(q − r) + (Q/q)(r − p) + (R/r)(p − q) = 0. P/p = a + (p − 1)d/2, and similarly for Q/q and R/r. Multiply each by the bracket and add: the a-terms give a[(q − r) + (r − p) + (p − q)] = 0 and the d-terms give (d/2)[(p − 1)(q − r) + (q − 1)(r − p) + (r − 1)(p − q)], which expands to zero because both the p, q, r products and the −1 terms cancel.

For GPs. If a, b, c are in GP, then b² = ac (b is the geometric mean); conversely b² = ac with all nonzero implies a GP. If a, b, c are in GP and x, y, z are their logarithms to any base, then 2y = x + z: log b² = log(ac) gives 2 log b = log a + log c, so the logarithms are in AP. This links the two progressions: the logarithms of a GP form an AP.

In every proof, state what is given in symbols, state what is to be proved in symbols, and manipulate one side until it becomes the other. Never substitute numbers for the letters to "check" as a substitute for the proof; a check is welcome after the proof, not instead of it.

The whole chapter rests on four formulas and two tests. For an AP with first term a and common difference d: a_n = a + (n − 1)d and S_n = n/2[2a + (n − 1)d] = n/2(a + l). For a GP with first term a and common ratio r: a_n = arⁿ⁻¹. The tests: constant difference for an AP, constant ratio for a GP. Everything else is recognising which of these to use and doing the algebra cleanly.

One-mark questions ask for the common difference or ratio of a given list, the next term, the nth term formula of a simple AP, whether a given list is an AP, or the 10th term of a stated progression. Answer in one or two lines, but show the subtraction or division that gives d or r.

Two-mark questions ask for a specific term, the number of terms of a finite AP, a value of x making three expressions an AP, the sum of a short AP, or a GP's nth term with a fractional or negative ratio. State the formula, substitute, compute.

Four-mark questions are the word problems: penalties, prizes, savings, logs, salaries, trees; the two-condition problems giving a pair of linear equations in a and d; the sum problems that lead to a quadratic in n; and the short proofs. The marking scheme gives credit for identifying a, d (or r) and n from the story, for writing the formula before substituting, for correct arithmetic, and for the final statement with units. A negative or fractional n must be rejected in writing.

Common errors to avoid. Using n instead of n − 1 in the nth term. Taking d as the previous term minus the next, so the sign is wrong. Forgetting that the sum formula has n/2 outside and 2a inside. Confusing S_n with a_n. Treating a list with a constant second difference (the squares) as an AP. Assuming a GP's ratio is positive when the signs alternate. In a problem giving two sums, forgetting to divide out n/2 before forming the linear equations. Writing (√2)ⁿ⁻¹ = 64 and then guessing n instead of writing 64 = (√2)¹².

Approach to any problem. (1) Write the first three or four terms from the story. (2) Decide AP or GP by testing. (3) Identify a, d or r, and which of n, a_n, S_n is given and which is asked. (4) Write the formula in symbols. (5) Substitute and solve; if a quadratic in n appears, factorise and reject impossible roots. (6) Check the answer against the story: does the 16th row really have 5 logs, does 4 + … really give 78? (7) State the answer with units.

What comes next: the Intermediate course adds the sum of n terms of a GP, the sum to infinity when |r| < 1, harmonic progressions, and the sums of squares and cubes of the first n natural numbers. The nth-term reasoning of this chapter is the foundation of all of that.