Sequences and Progressions — Detailed Notes
A sequence is an ordered list of numbers; each number is a term. We write the terms as t₁, t₂, t₃, … (the subscript is the position). Sequences can be finite or infinite. Familiar ones: natural (1,2,3,…), odd (1,3,5,…), triangular (1,3,6,10,…), square (1,4,9,16,…).
8.2–8.3 Explicit & recursive rules
• An explicit rule gives the n-th term directly from n, e.g. uₙ = 2n − 1 (odd numbers). It lets you find any term (the 53rd, 300th…) without the earlier ones, and to check if a number is a term (solve uₙ = that number; n must be a natural number). • A recursive rule gives a term from earlier terms, e.g. t₁ = 1, tₙ = tₙ₋₁ + 3. The famous Virahāṅka–Fibonacci sequence uses the previous two: Vₙ = Vₙ₋₁ + Vₙ₋₂ → 1, 2, 3, 5, 8, 13, 21, …
8.4 Arithmetic Progressions (AP)
AP: a, a+d, a+2d, … • n-th term tₙ = a + (n − 1)d
A list in which the difference between consecutive terms is a constant common difference d. Recursive form: t₁ = a, tₙ = tₙ₋₁ + d. Plotting (position, term) for an AP gives points on a straight line (slope d).
8.5 Sum of the first n natural numbers
1 + 2 + 3 + … + n = n(n + 1) / 2
Found by writing the sum forwards and backwards and adding (each pair totals n + 1). This is also the n-th triangular number, tₙ = n(n+1)/2. (First recorded by Āryabhaṭa.)
8.6 Geometric Progressions (GP)
GP: a, ar, ar², … • n-th term tₙ = a·rⁿ⁻¹
A list in which each term is the previous one times a constant common ratio r (so t₂/t₁ = t₃/t₂ = … = r). Recursive form: t₁ = a, tₙ = r·tₙ₋₁. Plotting a GP gives points on a curve, not a line. Fractals such as the Sierpiński triangle give GPs — number of black triangles = 3ⁿ (growing fast) while the black area = (3/4)ⁿ (shrinking to 0).
Key Terms
sequence, term, finite/infinite, explicit rule, recursive rule, Virahāṅka–Fibonacci, arithmetic progression, common difference, sum of first n naturals, triangular number, geometric progression, common ratio, fractal.
End-of-Chapter Exercises — Full Solutions
Complete worked solutions to all 15 end-of-chapter problems (starred * = HOTS), plus key in-text exercise answers.
1Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
t₁₁ = a + 10d = 38 and t₁₆ = a + 15d = 73 ⇒ 5d = 35 ⇒ d = 7, a = 38 − 70 = −32. t₃₁ = a + 30d = −32 + 210 = 178.
2Determine the AP whose 3rd term is 16 and whose 7th term exceeds the 5th term by 12.
a + 2d = 16; t₇ − t₅ = 2d = 12 ⇒ d = 6 ⇒ a = 16 − 12 = 4. The AP is 4, 10, 16, 22, 28, …
3(*) How many three-digit numbers are divisible by 7?
The three-digit multiples of 7 form an AP with first term 105 and last term 994, common difference 7. n = (994 − 105)/7 + 1 = 889/7 + 1 = 127 + 1 = 128.
4(*) How many multiples of 4 lie between 10 and 250?
The multiples of 4 here run from 12 to 248 (AP, d = 4). n = (248 − 12)/4 + 1 = 236/4 + 1 = 59 + 1 = 60.
5(*) Find a GP whose first two terms sum to −4 and whose 5th term is 4 times the 3rd term.
t₅ = 4·t₃ ⇒ ar⁴ = 4ar² ⇒ r² = 4 ⇒ r = ±2. And a + ar = a(1 + r) = −4. If r = 2: a(3) = −4 ⇒ a = −4/3 → GP −4/3, −8/3, −16/3, … If r = −2: a(−1) = −4 ⇒ a = 4 → GP 4, −8, 16, −32, … (both are valid).
6(*) Find all ways to write 100 as a sum of consecutive natural numbers.
Writing k consecutive numbers from m gives k(2m + k − 1) = 200. The integer solutions are: 18 + 19 + 20 + 21 + 22 = 100 (5 terms) and 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100 (8 terms).
7(*) Bacteria double every hour, starting with 30. How many are present at the end of the 2nd hour, 4th hour and n-th hour?
This is a GP with initial 30 and ratio 2. After the 2nd hour: 30 × 2² = 120; after the 4th hour: 30 × 2⁴ = 480; after the n-th hour: 30 × 2ⁿ.
8(*) In an AP, t₄ + t₈ = 24 and t₆ + t₁₀ = 44. Find the first three terms.
t₄ + t₈ = 2a + 10d = 24 ⇒ a + 5d = 12. t₆ + t₁₀ = 2a + 14d = 44 ⇒ a + 7d = 22. Subtracting: 2d = 10 ⇒ d = 5, a = 12 − 25 = −13. First three terms: −13, −8, −3.
9(*) Find the smallest n for which the sum of the first n natural numbers exceeds 1000.
Need n(n+1)/2 > 1000 ⇒ n(n+1) > 2000. For n = 44: 44×45 = 1980 < 2000; for n = 45: 45×46 = 2070 > 2000. So the smallest value is n = 45.
10(*) Which term of the GP 2, 8, 32, … is 131072? Give the explicit and recursive formulas.
Here a = 2, r = 4, so tₙ = 2·4ⁿ⁻¹. Set 2·4ⁿ⁻¹ = 131072 ⇒ 4ⁿ⁻¹ = 65536 = 4⁸ ⇒ n − 1 = 8 ⇒ n = 9 (the 9th term). Explicit: tₙ = 2·4ⁿ⁻¹; recursive: t₁ = 2, tₙ = 4·tₙ₋₁.
11(*) The sum of the first three terms of a GP is 13/12 and their product is −1. Find the common ratio and the terms.
Take the terms as a/r, a, ar. Product = a³ = −1 ⇒ a = −1. Sum: −1(1/r + 1 + r) = 13/12 ⇒ r + 1/r = −25/12 ⇒ 12r² + 25r + 12 = 0 ⇒ r = −3/4 or −4/3. The terms are 4/3, −1, 3/4 (with common ratio −3/4; the ratio −4/3 just reverses the order).
12(*) If the 4th, 10th and 16th terms of a GP are x, y, z, prove that x, y, z are in GP.
x = ar³, y = ar⁹, z = ar¹⁵. Then y² = (ar⁹)² = a²r¹⁸ and xz = (ar³)(ar¹⁵) = a²r¹⁸. Since y² = xz, the numbers x, y, z are in GP (consecutive ratios equal).
13(*) The sum of the first three terms of a GP is 26 and the sum of their squares is 364. Find the terms.
Let the terms be a, ar, ar². Using a(1+r+r²) = 26 and a²(1+r²+r⁴) = 364, and 1+r²+r⁴ = (1+r+r²)(1−r+r²), we get ar = 6 and a(1+r²) = 20 ⇒ 3r² − 10r + 3 = 0 ⇒ r = 3 or 1/3. With r = 3, a = 2: the terms are 2, 6, 18 (r = 1/3 gives the same set reversed).
14(*) P₁ = 1, P₂ = 2, and Pn = P₁ + P₂ + … + Pₙ₋₁ + 1 for n > 2. Find P₁…P₈; give a simpler recursive and an explicit formula.
P₃ = 1+2+1 = 4, P₄ = 1+2+4+1 = 8, P₅ = 16, P₆ = 32, P₇ = 64, P₈ = 128 → 1, 2, 4, 8, 16, 32, 64, 128. Simpler recursive rule: Pₙ = 2·Pₙ₋₁ (each term doubles, for n ≥ 3). Explicit: Pₙ = 2ⁿ⁻¹.
15(*) W₁ = 1, W₂ = 2, and Wn = W₁ + W₂ + … + Wₙ₋₂ + 2 for n > 2. Find W₁…W₈. Do you recognise it?
W₃ = 1+2 = 3, W₄ = 1+2+2 = 5, W₅ = 1+2+3+2 = 8, W₆ = 1+2+3+5+2 = 13, W₇ = 21, W₈ = 34 → 1, 2, 3, 5, 8, 13, 21, 34. This is the Virahāṅka–Fibonacci sequence (each term is the sum of the previous two).
IXKey in-text exercise answers (Sets 8.2 & 8.3).
Set 8.2 Q1 — AP 3,8,13,18,… (a=3, d=5): t₁₀ = 3+9×5 = 48, t₂₆ = 3+25×5 = 128. Set 8.2 Q5 — two-digit multiples of 3 run 12–99 (30 terms); sum = (12+99)×30/2 = 1665. Set 8.2 Q7 — 1+2+…+25 = 25×26/2 = 325 marbles. Set 8.3 Q1 — GP, r=2, t₈ = 192 ⇒ t₁₂ = 192×2⁴ = 3072. Set 8.3 Q2 — GP 5,25,125,… (a=5,r=5): tₙ = 5ⁿ, t₁₀ = 5¹⁰.