Chapter 11 — Progressions Optional

The main exercises of the chapter ask for one thing at a time: a term, a sum, a common difference. The optional exercise asks for two or three things at once, or hides the progression inside a picture. The formulas do not change — a_n = a + (n − 1)d, S_n = n/2[2a + (n − 1)d], a_n = arⁿ⁻¹ — but the problems require you to write two or more equations from the story, to solve a quadratic in n, or to see that a picture of rungs or steps is an AP in disguise.

Four families cover almost everything. Family 1 — two conditions, two unknowns. "The sum of the 4th and 8th terms is 24, and the sum of the 6th and 10th terms is 44." Each condition is a linear equation in a and d; solve the pair. Family 2 — a sum fixed, n unknown. "How many terms make the sum 636?" The sum formula becomes a quadratic in n; factorise and reject the impossible root. Family 3 — a picture that is an AP. Rungs of a ladder, steps of a staircase, houses numbered 1 to 49, circles of increasing radius. Identify what quantity forms the AP, then apply the formula. Family 4 — reasoning and proof. "Show that the (m + n)th term is zero", "is this sequence an AP?", "which term is the first negative one?" These are answered in letters, not numbers.

The examination rewards a fixed presentation. State the unknowns (a, d, n). Write each condition as an equation with a one-line reason. Solve, showing the elimination. Compute what is asked. Reject impossible values in words. Check against the story. This presentation is worth as much as the arithmetic; a correct answer with no equations shown earns little.

A caution about signs: many optional problems have negative common differences (a ladder narrowing upward, a decreasing sequence) and the sum can be negative or a quadratic in n can have both roots positive. Read "decreasing" as d < 0, "first negative term" as a_n < 0, and do not assume the larger root of a quadratic is the answer. The chapter now takes the four families in order, with GPs at the end.

When a problem states two facts about the terms of an AP, each fact becomes a linear equation in the first term a and the common difference d, and the two together determine the AP.

Example 1. The sum of the 4th and 8th terms of an AP is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms. Write a₄ + a₈ = (a + 3d) + (a + 7d) = 2a + 10d = 24, so a + 5d = 12. And a₆ + a₁₀ = (a + 5d) + (a + 9d) = 2a + 14d = 44, so a + 7d = 22. Subtracting, 2d = 10, d = 5, and a = 12 − 25 = −13. The AP is −13, −8, −3, ….

Example 2. The 17th term of an AP exceeds its 10th term by 7. Find d. (a + 16d) − (a + 9d) = 7d = 7, so d = 1. Note that a cancels; one condition suffices because only d is asked.

Example 3. If the 3rd and 9th terms of an AP are 4 and −8, which term is zero? a + 2d = 4 and a + 8d = −8. Subtracting, 6d = −12, d = −2, a = 8. Then 8 + (n − 1)(−2) = 0 gives n − 1 = 4, n = 5. The 5th term is zero.

Example 4. The sum of the first 6 terms is 42, and the ratio of the 10th term to the 30th term is 1 : 3. Find the first term and the 13th term. S₆ = 3(2a + 5d) = 42, so 2a + 5d = 14. And (a + 9d)/(a + 29d) = 1/3 gives 3a + 27d = a + 29d, so 2a = 2d, a = d. Then 2d + 5d = 14, d = 2, a = 2. The 13th term is 2 + 24 = 26.

Example 5. The 8th term of an AP is half its 2nd term, and the 11th term exceeds one-third of the 4th term by 1. Find the 15th term. a + 7d = (a + d)/2 gives 2a + 14d = a + d, so a = −13d. And a + 10d = (a + 3d)/3 + 1 gives 3a + 30d = a + 3d + 3, so 2a + 27d = 3. Substituting, −26d + 27d = 3, d = 3, a = −39. The 15th term is −39 + 42 = 3.

The method never varies: express every term mentioned as a + (k − 1)d, form the equations, and eliminate. The most frequent error is an arithmetic slip in the coefficient (writing a + 8d for the 8th term instead of a + 7d). Say the rule aloud as you write: the kth term has (k − 1) d's.

When the sum of an unknown number of terms is given, the sum formula becomes a quadratic in n. The optional exercise's standard case is: how many terms of the AP 9, 17, 25, … must be taken to give a sum of 636?

Here a = 9, d = 8, S_n = 636. So n/2[18 + 8(n − 1)] = 636, that is, n(8n + 10) = 1272, 8n² + 10n − 1272 = 0, 4n² + 5n − 636 = 0. To factorise, find two numbers with product 4 × (−636) = −2544 and sum 5: they are 53 and −48. So 4n² + 53n − 48n − 636 = n(4n + 53) − 12(4n + 53) = (4n + 53)(n − 12) = 0. Thus n = 12 (the root −53/4 is rejected). Check: the 12th term is 9 + 88 = 97, and S = 6(9 + 97) = 636.

When the factorisation is hard to see, use the formula: n = [−5 ± √(25 + 10176)]/8 = [−5 ± √10201]/8 = (−5 ± 101)/8, giving 12 or −13.25. Recognising 10201 = 101² is the only difficulty.

Example 2. How many terms of 18, 16, 14, … give a sum of 78? a = 18, d = −2: n/2[36 − 2(n − 1)] = 78, n(38 − 2n) = 156, n² − 19n + 78 = 0, (n − 6)(n − 13) = 0. Both roots are positive integers. n = 6: 18 + 16 + 14 + 12 + 10 + 8 = 78. n = 13: terms 7 to 13 are 6, 4, 2, 0, −2, −4, −6, which add to 0, so the sum is still 78. Both answers are correct and both must be stated.

Example 3. The first term of an AP is 5, the last term is 45 and the sum is 400. Find n and d. Here the short form applies: 400 = n/2(5 + 45) = 25n, so n = 16. Then 45 = 5 + 15d, d = 8/3. No quadratic is needed when a, l and S are given.

Example 4. The first and last terms of an AP are 17 and 350, and d = 9. How many terms are there and what is their sum? 350 = 17 + 9(n − 1), n − 1 = 37, n = 38. S = 19(17 + 350) = 19 × 367 = 6973.

Example 5 — sum to n terms given as a function. If S_n = 3n² + 5n, find the AP and its 20th term. a₁ = S₁ = 8; a₂ = S₂ − S₁ = 22 − 8 = 14; d = 6; a₂₀ = 8 + 114 = 122. Alternatively a_n = S_n − S_n−1 = 6n + 2.

The rule for two positive roots: both are valid only if the terms between them sum to zero, which requires the AP to pass through zero symmetrically. If the second root gives a negative number of objects (as in the logs problem), or the story forbids negative terms, reject it and say why.

Some problems give the sums of two separate blocks of terms, or the ratio of two sums, and ask for the AP or for another sum. The tool is that the sum of terms from the (p + 1)th to the qth equals S_q − S_p.

Example 1. The sum of the first ten terms of an AP is 210, and the sum of its next ten terms (the 11th to the 20th) is 610. Find the AP. S₁₀ = 5(2a + 9d) = 210, so 2a + 9d = 42. The next ten terms have sum S₂₀ − S₁₀ = 610, so S₂₀ = 820 and 10(2a + 19d) = 820, so 2a + 19d = 82. Subtracting, 10d = 40, d = 4, and 2a = 42 − 36 = 6, a = 3. The AP is 3, 7, 11, 15, ….

Example 2. The sum of the first n terms of an AP is 4n − n². Find the first term, the sum of the first two terms, the second term, and the nth term. S₁ = 3 = a. S₂ = 8 − 4 = 4, so a₂ = 4 − 3 = 1 and d = −2. a_n = S_n − S_n−1 = (4n − n²) − [4(n − 1) − (n − 1)²] = 4 − 2n + 1 = 5 − 2n. Check: a₁ = 3, a₂ = 1, a₃ = −1, a₁₀ = −15.

Example 3 — ratio of sums. The ratio of the sums of the first m and first n terms of an AP is m² : n². Show that the ratio of the mth and nth terms is (2m − 1) : (2n − 1). S_m/S_n = [m(2a + (m − 1)d)]/[n(2a + (n − 1)d)] = m²/n², so [2a + (m − 1)d]/[2a + (n − 1)d] = m/n, so 2an + (m − 1)nd = 2am + (n − 1)md, so 2a(n − m) = d(nm − m − nm + n) = d(n − m), hence d = 2a. Then a_m/a_n = [a + (m − 1)2a]/[a + (n − 1)2a] = (2m − 1)/(2n − 1).

Example 4 — sums of two APs in a ratio. The sums of the first n terms of two APs are in the ratio (7n + 1) : (4n + 27). Find the ratio of their 9th terms. The nth term is S_n − S_n−1, but a cleaner route uses the fact that a_k = [2a + (k − 1)d]/2 × 2/2, and S_2k−1/(2k − 1) is exactly the kth term's form [2a + (2k − 2)d]/2 = a + (k − 1)d. So the ratio of the 9th terms equals the ratio of the sums with n = 2 × 9 − 1 = 17: (7 × 17 + 1)/(4 × 17 + 27) = 120/95 = 24/19.

Example 5. If S_n denotes the sum of the first n terms, show that S₃₀ = 3(S₂₀ − S₁₀). S₂₀ − S₁₀ = 10(2a + 19d) − 5(2a + 9d) = 10a + 145d. And S₃₀ = 15(2a + 29d) = 30a + 435d = 3(10a + 145d). Proved.

These problems test whether S_n is understood as a function of n; once that is clear, differences and ratios of sums are ordinary algebra.

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 is the total length of wood required for the rungs?

This is Family 3: a picture that is an AP. Two separate APs are hidden here. The positions of the rungs are 25 cm apart over a span of 250 cm, so the number of gaps is 250/25 = 10 and the number of rungs is 11. The lengths of the rungs decrease uniformly, so they form an AP with first term 45, last term 25 and 11 terms. The total wood is S₁₁ = 11/2 (45 + 25) = 11 × 35 = 385 cm.

Notice what "decrease uniformly" means: the same amount is lost from rung to rung, which is exactly the AP condition. The common difference is (25 − 45)/10 = −2 cm per rung, so the lengths are 45, 43, 41, …, 25. It is not needed for the sum, since the short form n/2(a + l) suffices, but it lets you list the rungs if asked.

The trap is the count of rungs. A span of 250 cm with rungs 25 cm apart has 10 intervals but 11 rungs — the fencepost principle. Students who write n = 10 get 10/2(45 + 25) = 350 and lose the mark. Draw the ladder, mark the bottom rung as rung 1 at height 0 and the top rung at height 250, and count.

Variants of the same structure: a tapering flagpole built from cylindrical sections whose diameters decrease uniformly; a stack of books whose widths decrease by a fixed amount; a pyramid of cans with rows decreasing by one; a set of shelves with lengths decreasing by 5 cm. In each, first find n from the spacing or from the count, then find a and l from the extreme sizes, then sum.

A second variant supplies a and d and asks for the top: a ladder of 15 rungs whose bottom rung is 60 cm and each rung is 2.5 cm shorter than the one below has top rung 60 − 14 × 2.5 = 25 cm and total wood 15/2 (60 + 25) = 637.5 cm.

A third variant reverses the question: if 385 cm of wood made 11 rungs whose lengths decrease uniformly from 45 cm, find the top rung: 385 = 11/2 (45 + l), 45 + l = 70, l = 25 cm.

For the examination, write the three steps separately — number of rungs, the AP of lengths, the sum — and give the answer in the unit of the question (cm or m).

The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding house x is equal to the sum of the numbers of the houses following it, and find x.

The houses before x are numbered 1, 2, …, x − 1, an AP with sum (x − 1)x/2. The houses after x are x + 1, …, 49, and their sum is the sum of 1 to 49 minus the sum of 1 to x, that is, 49 × 50/2 − x(x + 1)/2 = 1225 − x(x + 1)/2.

Setting these equal: (x − 1)x/2 = 1225 − x(x + 1)/2. Multiply by 2: x² − x = 2450 − x² − x, so 2x² = 2450, x² = 1225, x = 35. The negative root is rejected. Check: 1 + 2 + … + 34 = 34 × 35/2 = 595, and 36 + 37 + … + 49 = 1225 − 35 × 36/2 = 1225 − 630 = 595. Equal.

The elegance is in noticing that the x-terms cancel, leaving x² = (49 × 50)/2 = 1225, so x = √1225. In general, for houses numbered 1 to N, the balancing house satisfies x² = N(N + 1)/2, so a solution exists only when N(N + 1)/2 is a perfect square: for N = 8, x² = 36, x = 6 (1 + … + 5 = 15 = 7 + 8); for N = 49, x = 35; for N = 288, x = 204. For most N there is no such house.

A close relative is the problem of a number x between 1 and N for which the sum of the numbers before it equals the sum after it, or a cutting of a stack of numbered cards. The reasoning is the same, and the formula 1 + 2 + … + n = n(n + 1)/2 does all the work.

Another relative: find the sum of all houses except x, or find x so that the sum of numbers before x is twice the sum after. The latter gives x(x − 1)/2 = 2[1225 − x(x + 1)/2], so x² − x = 4900 − 2x² − 2x, 3x² + x − 4900 = 0, whose roots are not integers, so no such house exists — an acceptable examination answer when justified.

The examination expects: the two sums written with the formula, the equation, the cancellation, x = 35, and the check. The check is not optional here; it is the natural way to confirm the cancellation was done correctly.

A small terrace is at the end of a footpath. It has a flight of 15 steps, each of rise ¼ m and tread ½ m, and the terrace is 50 m wide. Each step is built by stacking concrete on the full width, so the first step is a cuboid ¼ m high, ½ m deep and 50 m wide; the second step is ½ m high (it sits on the ground beside the first, rising to the second level), ½ m deep and 50 m wide; and so on. Find the total volume of concrete.

The volume of the kth step is height × depth × width = (k/4) × (1/2) × 50 = 25k/4 m³. The volumes for k = 1, 2, …, 15 are 25/4, 50/4, 75/4, …, an AP with a = 25/4 and d = 25/4. The total is S₁₅ = 15/2 [2 × 25/4 + 14 × 25/4] = 15/2 × 25/4 × 16 = 15 × 25 × 2 = 750 m³.

A quicker route: the total is (25/4)(1 + 2 + … + 15) = (25/4)(120) = 750. Recognising a common factor and using n(n + 1)/2 is often the fastest path for a "sum of multiples" AP.

The key idea is that each step is a full cuboid down to the ground, not just the visible tread; the heights grow 1/4, 2/4, 3/4, … so the volumes grow in proportion. Draw the cross-section: a staircase profile whose successive columns have heights k/4 and equal widths 1/2, and the total area of the profile is (1/2) × (1/4)(1 + 2 + … + 15) = 15 m², which times the 50 m width is 750 m³.

Related solid stacks. A pyramid of cubes built in square layers 1, 4, 9, 16, 25 is not an AP (differences 3, 5, 7, 9) and its total 55 must be added directly or by the sum-of-squares formula n(n + 1)(2n + 1)/6 from the Intermediate course. A triangular wall of bricks with 1 brick in the top row, 2 in the next, …, 20 in the bottom row has 20 × 21/2 = 210 bricks, an AP. A conical heap made of layers whose radii increase uniformly is an AP in radius but a sum of squares in volume — distinguish the linear from the quadratic quantity before summing.

Circles of increasing radius. A design of concentric circles with radii 1, 2, 3, …, 10 cm has total circumference 2π(1 + 2 + … + 10) = 2π × 55 = 110π cm; the circumferences form an AP with d = 2π. The total area π(1 + 4 + 9 + … + 100) = 385π cm² is a sum of squares, not an AP.

Problems in which terms are inserted between two numbers, or in which the terms themselves are unknown and must be found from their sum and another condition, are solved by the arithmetic mean and the symmetric choice of terms.

Insertion. Insert 6 numbers between 3 and 24 so that the resulting sequence is an AP. There are 8 terms, a = 3, a₈ = 24, so 3 + 7d = 24, d = 3, and the numbers are 6, 9, 12, 15, 18, 21. In general, inserting k terms between p and q gives d = (q − p)/(k + 1).

Unknown terms from sum and product. Find three numbers in AP whose sum is 21 and product is 231. Take a − d, a, a + d: 3a = 21, a = 7; 7(49 − d²) = 231, 49 − d² = 33, d² = 16, d = ±4. The numbers are 3, 7, 11 (or 11, 7, 3).

Unknown terms from sum and sum of squares. Four numbers in AP have sum 32 and the ratio of the product of the extremes to the product of the means is 7 : 15. Take a − 3d, a − d, a + d, a + 3d: 4a = 32, a = 8. (a − 3d)(a + 3d)/[(a − d)(a + d)] = (64 − 9d²)/(64 − d²) = 7/15, so 960 − 135d² = 448 − 7d², 512 = 128d², d² = 4, d = ±2. The numbers are 2, 6, 10, 14.

Angles of a polygon. The interior angles of a polygon are in AP with smallest angle 120° and common difference 5°. Find the number of sides. The sum of interior angles is (n − 2) × 180, and also n/2[240 + 5(n − 1)]. So (n − 2)180 = n(235 + 5n)/2, 360n − 720 = 235n + 5n², 5n² − 125n + 720 = 0, n² − 25n + 144 = 0, (n − 9)(n − 16) = 0. If n = 16, the largest angle is 120 + 75 = 195° > 180°, impossible for a convex polygon, so n = 9.

Middle term problems. Find the middle term of the AP 6, 13, 20, …, 216. Number of terms: 6 + 7(n − 1) = 216, n − 1 = 30, n = 31. The middle term is the 16th: 6 + 15 × 7 = 111, which is also (6 + 216)/2. If the number of terms were even, say 3, 7, …, 79 (n = 20), the two middle terms are the 10th and 11th, 39 and 43, and their mean is 41 = (3 + 79)/2.

Splitting a number into AP parts. Divide 32 into four parts in AP so that the ratio of the product of the extremes to the product of the means is 7 : 15 — this is the four-term problem above. Divide 24 into three parts in AP with product 440: a = 8, 8(64 − d²) = 440, d² = 9, parts 5, 8, 11.

The symmetric choices work because the sum immediately gives a, and a second condition — product, sum of squares, ratio — gives d² and hence d. Both signs of d give the same set of numbers in opposite order, so state the set once.

The GP problems of the optional exercise ask for the terms when their product or the ratio of two of them is given, or for a term far along the sequence when two others are known. The tools are the symmetric choice a/r, a, ar for three terms, the nth-term formula, and division of one term equation by another to eliminate a.

Three terms from product and sum. Find three numbers in GP whose sum is 14 and product is 64. Take a/r, a, ar: the product is a³ = 64, a = 4. The sum is 4/r + 4 + 4r = 14, so 4/r + 4r = 10, 4 + 4r² = 10r, 2r² − 5r + 2 = 0, (2r − 1)(r − 2) = 0, r = 2 or 1/2. The numbers are 2, 4, 8 (either order).

Terms from two given terms. The 5th term of a GP is 81 and the 2nd term is 3. Find the GP. ar⁴ = 81 and ar = 3. Dividing, r³ = 27, r = 3, a = 1. The GP is 1, 3, 9, 27, 81, …. If instead the 3rd term is 24 and the 6th is 192: r³ = 8, r = 2, a = 6.

Finding n. The first term of a GP is 5 and the common ratio 2. Which term is 1280? 5 × 2ⁿ⁻¹ = 1280, 2ⁿ⁻¹ = 256 = 2⁸, n = 9. If a = √3 and r = √3, which term is 27√3? √3 × (√3)ⁿ⁻¹ = (√3)ⁿ = 27√3 = 3³ × 3^1/2 = 3^7/2 = (√3)⁷, so n = 7.

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. In powers of 2: 2²ⁿ⁻¹ = 2¹¹⁻ⁿ, so 2n − 1 = 11 − n, n = 4. Both 4th terms are 128.

Product of terms equidistant from the ends. In a finite GP a, ar, …, arⁿ⁻¹, the product of the kth term from the start and the kth term from the end is ar^k−1 × ar^n−k = a²rⁿ⁻¹ = a × l, independent of k. This is the GP counterpart of a_k + a_n−k+1 = a + l for an AP.

Geometric mean and inserting terms. The geometric mean of 4 and 9 is √36 = 6, and 4, 6, 9 is a GP with r = 3/2. To insert k terms between p and q in GP, q = pr^k+1, so r = (q/p)^1/(k+1). Insert 3 terms between 1 and 256: r⁴ = 256, r = 4, terms 4, 16, 64. Insert 2 terms between 1 and 27: r³ = 27, r = 3, terms 3, 9.

Compound growth. A sum grows to 1.21 times itself in 2 years at compound interest; after 5 years it is (1.1)⁵ = 1.61051 times itself. A machine losing 10% of its value each year is worth V(0.9)ⁿ after n years, and it falls below half its value first when 0.9ⁿ < 0.5, which happens at n = 7 (0.9⁷ ≈ 0.478). Such questions test the exponent, and the working must show the powers.

The reasoning questions of the optional exercise are short proofs that follow from writing the general term as a + (n − 1)d and manipulating exactly.

Theorem 1. If m times the mth term of an AP equals n times the nth term, the (m + n)th term is zero. Proof: m[a + (m − 1)d] = n[a + (n − 1)d]. Rearranging, a(m − n) + d[m² − m − n² + n] = 0, that is, a(m − n) + d(m − n)(m + n − 1) = 0. Since m ≠ n, divide by (m − n): a + (m + n − 1)d = 0, which is a_m+n. Proved.

Theorem 2. If the pth, qth and rth terms of an AP are x, y, z respectively, then x(q − r) + y(r − p) + z(p − q) = 0. Proof: x = a + (p − 1)d, and similarly for y and z. Then the left side is a[(q − r) + (r − p) + (p − q)] + d[(p − 1)(q − r) + (q − 1)(r − p) + (r − 1)(p − q)]. The first bracket is 0. The second: pq − pr + qr − pq + rp − rq = 0 from the products, and −(q − r) − (r − p) − (p − q) = 0 from the −1 terms. So the whole is 0.

Theorem 3. If S_p = q and S_q = p, then S_p+q = −(p + q). Proof: p/2[2a + (p − 1)d] = q and q/2[2a + (q − 1)d] = p. Multiply out: 2ap + p(p − 1)d = 2q and 2aq + q(q − 1)d = 2p. Subtract: 2a(p − q) + d[p² − p − q² + q] = 2(q − p), so (p − q)[2a + (p + q − 1)d] = −2(p − q), so 2a + (p + q − 1)d = −2. Then S_p+q = (p + q)/2 × (−2) = −(p + q).

Theorem 4. The sum of the first n odd natural numbers is n². Proof: 1 + 3 + … + (2n − 1) is an AP with a = 1, d = 2: S = n/2[2 + 2(n − 1)] = n/2 × 2n = n².

Theorem 5. If a, b, c are in GP, then a² + b², ab + bc, b² + c² are in GP. Given b² = ac. (ab + bc)² = b²(a + c)² = ac(a + c)² and (a² + b²)(b² + c²) = (a² + ac)(ac + c²) = a(a + c) × c(a + c) = ac(a + c)². Equal, so the middle squared equals the product of the outer two, and the three are in GP.

Theorem 6. Logarithms of a GP form an AP. If a, ar, ar², … is a GP with positive terms, then log(arⁿ⁻¹) = log a + (n − 1) log r, which is the nth term of an AP with first term log a and difference log r. Conversely, if x, y, z are in AP then 10^x, 10^y, 10^z are in GP.

The pattern of every proof: write the given in symbols using the general term or sum; write what is to be shown in symbols; transform the given by adding, subtracting or dividing equations until the target appears. Number the equations; say "subtracting (2) from (1)"; and finish with the statement proved. The examiner is grading the logic, not the final line.

A group of optional questions gives a sequence by a rule or by its first few terms and asks: is it an AP? is it a GP? what is a_n? which term first satisfies a condition? These need clear tests and clean algebra.

Sequences defined by a formula. a_n = 3 + 2n gives 5, 7, 9, 11, …: since a_n is linear in n, this is an AP with d = 2 and a = 5. a_n = 1 + n + n² gives 3, 7, 13, 21, …: the differences 4, 6, 8 are not constant, so not an AP, and the ratios are not constant, so not a GP. a_n = 2 × 3ⁿ gives 6, 18, 54, …: since a_n is a constant times rⁿ, this is a GP with r = 3 and a = 6. a_n = n² gives 1, 4, 9, 16: neither. The rule: a linear formula in n is an AP, an exponential formula is a GP, anything else is neither.

The first negative term. Which term of 20, 19¼, 18½, 17¾, … is the first negative term? a = 20, d = −3/4. a_n < 0 means 20 − (3/4)(n − 1) < 0, so (n − 1) > 80/3 = 26.67, so n − 1 ≥ 27, n ≥ 28. The 28th term, 20 − 27 × 3/4 = −1/4, is the first negative one; the 27th is 20 − 19.5 = 1/2 > 0.

Which term exceeds a bound. Which term of 5, 10, 20, … first exceeds 1000? 5 × 2ⁿ⁻¹ > 1000, 2ⁿ⁻¹ > 200; since 2⁷ = 128 and 2⁸ = 256, n − 1 = 8, n = 9. The 9th term is 1280.

Sequences that look like progressions but are not. 1, 1, 2, 3, 5, 8: each term is the sum of the two before; differences 0, 1, 1, 2, 3 — neither AP nor GP. 2, 3, 5, 7, 11: primes, no rule of either kind. 1, 3, 6, 10, 15: triangular numbers, differences 2, 3, 4, 5 — the differences form an AP but the sequence itself does not. A sequence can be an AP without looking like one: a, a, a, a is an AP with d = 0 and a GP with r = 1; 1/2, 1/4, 0, −1/4 is an AP with d = −1/4.

An AP from a table. The table of fares 10, 18, 26, 34 for 1, 2, 3, 4 km is an AP with d = 8; the fare for 15 km is 10 + 14 × 8 = 122. The heights of a plant measured weekly as 12, 15, 18, 21 cm form an AP and predict 12 + 3(n − 1) at week n. The amounts of a fixed deposit at compound interest 1000, 1080, 1166.40 form a GP and predict 1000(1.08)ⁿ. Modelling means naming the progression, finding a and d or r, and then extrapolating with the nth-term formula — with the caveat that a real plant does not grow linearly for ever.

Number of terms of a finite AP with fractional d. How many terms are in 3, 3½, 4, …, 10? d = 1/2, 3 + (n − 1)/2 = 10, n − 1 = 14, n = 15. Always compute d exactly as a fraction and keep n an integer.

This topic solves a set of problems in the style of the optional exercise and the harder examination questions, each with the full working, to show the method in action.

Problem 1. A sum of ₹1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years. Interest each year is 80, so the cumulative interests are 80, 160, 240, … an AP with a = 80, d = 80. After 30 years: 80 + 29 × 80 = ₹2400. (Compare compound interest, which would give a GP of amounts.)

Problem 2. In a flower bed there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 in the last row. How many rows are there? a = 23, d = −2, l = 5: 23 − 2(n − 1) = 5, n − 1 = 9, n = 10 rows. Total plants: 10/2 (23 + 5) = 140.

Problem 3. The sum of the three numbers in AP is 12 and the sum of their cubes is 288. Find the numbers. a − d, a, a + d: 3a = 12, a = 4. (4 − d)³ + 64 + (4 + d)³ = 288. Expand: (64 − 48d + 12d² − d³) + 64 + (64 + 48d + 12d² + d³) = 192 + 24d² = 288, so d² = 4, d = ±2. Numbers 2, 4, 6.

Problem 4. The 10th term of an AP is 52 and the 16th term is 82. Find the 32nd term and the general term. a + 9d = 52, a + 15d = 82, so 6d = 30, d = 5, a = 7. a₃₂ = 7 + 155 = 162; a_n = 5n + 2.

Problem 5. Find the sum of all natural numbers between 100 and 500 that are divisible by 8. First 104, last 496, d = 8: 104 + 8(n − 1) = 496, n = 50. S = 25(104 + 496) = 15000.

Problem 6. A manufacturer of TV sets produced 600 in the third year and 700 in the seventh year, the production increasing uniformly. Find the production in the first year, the 10th year, and the total in the first 7 years. a + 2d = 600, a + 6d = 700, d = 25, a = 550. a₁₀ = 550 + 225 = 775. S₇ = 7/2 (1100 + 150) = 7 × 625 = 4375.

Problem 7. Find the sum of the integers from 1 to 100 that are divisible by 2 or 5. Divisible by 2: 2 + … + 100, 50 terms, sum 2550. By 5: 5 + … + 100, 20 terms, sum 1050. By both (i.e. by 10): 10 + … + 100, 10 terms, sum 550. Required sum = 2550 + 1050 − 550 = 3050.

Problem 8. The GP 3, 6, 12, … and the AP 3, 8, 13, … have equal 1st terms; find the smallest n for which the GP's nth term exceeds the AP's. GP: 3 × 2ⁿ⁻¹; AP: 5n − 2. n = 2: 6 vs 8; n = 3: 12 vs 13; n = 4: 24 vs 18. So n = 4.

Each problem repeats the same moves — identify, equate, solve, reject, check — which is why practice on the optional exercise raises the score on the whole paper.

The SSC paper's progression questions are predictable in form, and a short strategy handles them all.

Read for the type. "Increases by a fixed amount", "decreases uniformly", "each row has one less", "simple interest" — AP. "Doubles", "increases by 10% each year", "compound interest", "half the previous" — GP. "Squares", "triangular numbers", "each term is the sum of the previous two" — neither; say so with the test.

Read for the given and the asked. Write the letters: a = ?, d = ?, n = ?, a_n = ?, S_n = ?. Fill in what the story gives. If two terms are given, you will solve a pair of linear equations. If a sum and a, d are given, you will solve a quadratic in n. If a picture is given, decide which quantity is the AP (lengths, counts, volumes) and find n from the geometry before summing.

Write the formula before substituting. The marking scheme allots a mark to the formula. "S_n = n/2[2a + (n − 1)d]" written once, then "= 15/2[400 + 29 × 50]", then the number.

Solve carefully. When a quadratic in n appears, factorise by splitting the middle term, or use the formula if the discriminant is a large perfect square (10201 = 101²). Reject negative, fractional and story-impossible roots in words. When two positive integer roots survive, test whether the terms between them sum to zero; if they do, state both answers.

Check. Substitute the answer back: does 4 + … give 78, does the 16th row have 5 logs, is the 28th term the first negative one and the 27th still positive? The check takes thirty seconds and catches most errors.

Time. A one-mark question deserves one minute, a four-mark problem six to eight. Do not derive the sum formula unless asked; when asked, the pairing derivation must show the reversed sum, the n equal pairs and the division by 2.

Common losses. Using n instead of n − 1; a sign error in d for a decreasing AP; counting intervals as rungs; forgetting that S_n is the sum and a_n the term; leaving a GP question with r found as "a₁/a₂" upside down; writing 2ⁿ⁻¹ = 256 and guessing n = 9 without writing 256 = 2⁸; in inclusion–exclusion, forgetting to subtract the overlap; in a proof, substituting numbers for the letters instead of arguing generally.

What the optional exercise prepares you for. The Intermediate course begins with the sum of a GP, S_n = a(rⁿ − 1)/(r − 1), and the infinite sum a/(1 − r) for |r| < 1, then harmonic progressions and the sums of squares and cubes. Every one of these builds on the two moves practised here: writing the general term, and reasoning with sums as functions of n. A student who can do the ladder, the houses and the m·a_m = n·a_n proof without hesitation is ready.