Arithmetic progression
If every step of a staircase has the same height, the height at the third or tenth step is easy to find. A sequence that grows by the same difference is an arithmetic progression. Building on classes 9–10 : Since classes 9–10 we know sequences and the first AP formula; here we go on to sums and means.
If every term is obtained by adding the same difference d to the previous one, the sequence is an arithmetic progression (AP). The first term is a and the common difference is d. The arithmetic mean of two numbers is their average.
Rule 1: Term and sum — The nth term is Tₙ = a + (n − 1)d. The sum of the first n terms is Sₙ = n/2 × (2a + (n − 1)d) = n/2 × (a + l), where l is the last term.
In the figure : Bars of the AP 3, 7, 11, …; each next bar is 4 taller.
Rule 2: Useful properties — Three numbers in AP: m − d, m, m + d. Sum and term: Tₙ = Sₙ − Sₙ₋₁. If the number of terms is odd, Sₙ = n × (middle term).
How is the formula obtained? : Write S = a + (a + d) + … + l and again in reverse order; adding the two lines gives (a + l) in every pair, n pairs. So 2S = n(a + l).
- Easy : The 15th term of 4, 9, 14, …: 4 + 14 × 5 = 74.
- Medium : Sum of the first 25 terms of 7, 10, 13, …: 25/2 × (14 + 24 × 3) = 25/2 × 86 = 1075.
- Exam-type : T₅ = 17 and T₉ = 33: 4d = 16, d = 4, a = 1. S₂₀ = 20/2 × (2 + 19 × 4) = 10 × 78 = 780.
In Tₙ the difference d is multiplied by (n − 1), not n; the first term has no difference added.
Taking the number of terms as n = (l − a)/d. The correct value is n = (l − a)/d + 1.
When the number of terms is odd, the sum = number of terms × the middle term; for 15 terms the middle is the 8th. Memory clue : "Sum = count × average".
In "how many terms" questions n must be a positive whole number; reject the negative root.
CET-style example : Sum of all multiples of 7 from 100 to 999? (A) 70336 (B) 70000 (C) 64000 (D) 70700
Answer : (A). First 105, last 994; n = (994 − 105)/7 + 1 = 128; S = 128/2 × 1099 = 64 × 1099 = 70336. Time-saving method : Find the first and last multiple, then n and n/2 × (a + l).
Spot the difference
| Point | AP | GP |
|---|---|---|
| Growth | adds the same d | multiplies by the same r |
| Tₙ | a + (n − 1)d | a × r^(n−1) |
| Mean | average (a + b)/2 | √(ab) |
Practice questions
- 1Find the 10th term of 2, 5, 8, ….
- 2State the common difference of 9, 6, 3, ….
- 3How many terms are in 5, 9, 13, …, 45?
- 4Find the sum of the first 12 terms of 3, 7, 11, ….
- 5The 4th term of an AP is 14 and the 9th term is 34; find the first term.
- 6Which is the first negative term of 7, 5, 3, …?
- 7Find the sum of all two-digit multiples of 6.
- 8Three numbers in AP have sum 18 and product 162; find the largest.
Show answers (try them yourself first)
- 12 + 9 × 3 = 29.
- 26 − 9 = −3.
- 3(45 − 5)/4 + 1 = 11.
- 412/2 × (6 + 11 × 4) = 6 × 50 = 300.
- 55d = 20, d = 4; a = 14 − 12 = 2.
- 6पदे 7, 5, 3, 1, −1 ; पाचवे.
- 712 ते 96 : 15 पदे ; 15/2 × 108 = 810.
- 8मधली संख्या 6 ; 6(36 − d²) = 162, d = 3 ; संख्या 3, 6, 9 ; मोठी 9.
Geometric progression
Things that double in each round, like cell division: 1, 2, 4, 8, …, grow by the same multiple every time. Such a sequence is a geometric progression. Building on classes 9–10 : Since classes 9–10 we know indices and ratios; a GP uses them.
If every term is obtained by multiplying the previous one by the same ratio r, it is a geometric progression (GP). The geometric mean of two numbers is √(ab).
Rule 1: Term and sum — Tₙ = a rⁿ⁻¹. Sₙ = a (rⁿ − 1)/(r − 1) (r > 1) or a (1 − rⁿ)/(1 − r) (r < 1). If r = 1 then Sₙ = na.
In the figure : Bars of the GP 1, 2, 4, 8, 16; each next bar is double.
Rule 2: Useful properties — Three numbers in GP: m/r, m, mr; the square of the middle one = the product of the other two. To insert n geometric means between a and b take r = (b/a)^(1/(n+1)).
How is the formula obtained? : S = a + ar + … + arⁿ⁻¹. Multiplying both sides by r and subtracting cancels all the middle terms and leaves S(1 − r) = a(1 − rⁿ).
- Easy : The 6th term of 2, 6, 18, …: 2 × 3⁵ = 2 × 243 = 486.
- Medium : Sum of 5 terms of 3 + 6 + 12 + …: 3 × (2⁵ − 1)/(2 − 1) = 3 × 31 = 93.
- Exam-type : T₂ = 6 and T₅ = 48: r³ = 8, r = 2, a = 3. S₆ = 3 × (2⁶ − 1) = 3 × 63 = 189.
For a negative r the signs of the terms alternate: 3, −6, 12, −24.
Writing Tₙ = a rⁿ. The correct form is a rⁿ⁻¹, because the first term is not multiplied by r.
When the product of three numbers is given, the middle number is the cube root: 64 → 4. Memory clue : "Product = middle cubed".
When r < 1 the formula (rⁿ − 1)/(r − 1) works but gives negative fractions; use (1 − rⁿ)/(1 − r) to make fewer errors.
CET-style example : Three numbers in GP have sum 14 and product 64; the largest is? (A) 8 (B) 16 (C) 4 (D) 12
Answer : (A). The middle is 4; 4/r + 4 + 4r = 14 gives 2r² − 5r + 2 = 0, r = 2; the numbers are 2, 4, 8. Time-saving method : Test the options: 2 + 4 + 8 = 14 and the product is 64.
Spot the difference
| Value of r | Behaviour of terms | Formula for Sₙ |
|---|---|---|
| r > 1 | grow | a (rⁿ − 1)/(r − 1) |
| 0 < r < 1 | shrink | a (1 − rⁿ)/(1 − r) |
| r < 0 | signs alternate | the matching formula above |
Practice questions
- 1Find the 5th term of 3, 6, 12, ….
- 2State the common ratio of 100, 50, 25, ….
- 3Find the geometric mean of 5 and 20.
- 4Find the sum of 6 terms of 1 + 3 + 9 + ….
- 5Which term of 2, 4, 8, … is 256?
- 6In a GP of positive terms T₃ = 20 and T₅ = 80; find the first term.
- 7How many terms of 1 + 2 + 4 + … give the sum 255?
- 8Three numbers in GP have sum 39 and product 729; find the largest.
Show answers (try them yourself first)
- 13 × 2⁴ = 48.
- 250 ÷ 100 = 1/2.
- 3√(5 × 20) = √100 = 10.
- 4(3⁶ − 1)/2 = 728/2 = 364.
- 52^(n−1) = 128 ; 2 × 2⁷ = 256 ; आठवे.
- 6r² = 80/20 = 4, r = 2 ; a = 20/4 = 5.
- 7(2ⁿ − 1) = 255 ; 2ⁿ = 256 ; n = 8.
- 8मधली 9 ; 9/r + 9 + 9r = 39 वरून 3r² − 10r + 3 = 0, r = 3 ; संख्या 3, 9, 27 ; मोठी 27.
Infinite GP and recurring decimals
If we cut half of a strip, then half of what is left, and so on, the total length cut gets close to the strip’s length but never passes it. So the sum of infinitely many terms can still be a definite number. Building on classes 9–10 : Since classes 9–10 we know fractions and decimals; now we turn recurring decimals into fractions by a formula.
In a GP with |r| < 1 the terms become so small that the sum to infinity settles at one number. A recurring decimal is the sum of such a GP.
Rule 1: Sum to infinity — S∞ = a/(1 − r), only when |r| < 1. If |r| ≥ 1 there is no sum to infinity.
In the figure : The partial sums of 1 + 1/2 + 1/4 + … approach 2; the gap halves every time.
Rule 2: Recurring decimals — Treat the repeating part as a GP and take its sum: 0.3535… = 0.35 + 0.0035 + …; a = 0.35, r = 0.01; S = 0.35/0.99 = 35/99. Keep the whole-number part separate.
How is the formula obtained? : Sₙ = a(1 − rⁿ)/(1 − r). When |r| < 1, rⁿ tends to 0 as n grows; so Sₙ → a/(1 − r).
- Easy : 4 + 2 + 1 + …: a = 4, r = 1/2, S = 4/(1/2) = 8.
- Medium : 0.4545…: a = 0.45, r = 0.01, S = 0.45/0.99 = 45/99 = 5/11.
- Exam-type : 1.2333… = 1.2 + 0.0333… = 6/5 + 0.03/0.9 = 6/5 + 1/30 = 36/30 + 1/30 = 37/30.
Check |r| < 1 before finding a sum to infinity; otherwise the formula does not apply.
Treating 0.999… as smaller than 1. With a = 0.9, r = 0.1, S = 0.9/0.9 = 1; they are equal.
Pure recurring: repeating digits ÷ the same number of 9s (0.6363… = 63/99). Mixed: (all digits − non-repeating digits) ÷ (9s then 0s). Memory clue : "Nines for the repeat, zeros for the rest".
For negative r the denominator becomes larger: 1 − (−1/2) = 3/2, not 1/2.
CET-style example : 1 − 1/2 + 1/4 − 1/8 + … = ? (A) 2/3 (B) 2 (C) 1/2 (D) 4/3
Answer : (A). a = 1, r = −1/2; S = 1/(1 + 1/2) = 2/3. Time-saving method : Note the sign of r and subtract it inside 1 − r.
Spot the difference
| Type | Example | Fraction |
|---|---|---|
| Pure recurring | 0.7777… | 7/9 |
| Pure recurring (two digits) | 0.6363… | 63/99 = 7/11 |
| Mixed | 0.1666… | (16 − 1)/90 = 1/6 |
Practice questions
- 1Find the sum to infinity of 8 + 4 + 2 + ….
- 2Find the sum to infinity of the GP with a = 3, r = 1/10.
- 3Convert 0.5555… to a fraction.
- 4Convert 0.0303… to a fraction.
- 5The sum to infinity of a GP is 20 and r = 3/5; find the first term.
- 6The sum to infinity of a GP is 6 and its first term is 2; find r.
- 7Convert 3.1818… to a fraction.
- 8Find the sum of 1/9 + 1/27 + 1/81 + ….
Show answers (try them yourself first)
- 18/(1/2) = 16.
- 23/(9/10) = 10/3.
- 30.5/0.9 = 5/9.
- 4a = 0.03, r = 0.01 ; 0.03/0.99 = 3/99 = 1/33.
- 5a = 20 × (2/5) = 8.
- 66 = 2/(1 − r) ; 1 − r = 1/3 ; r = 2/3.
- 73 + 18/99 = 3 + 2/11 = 35/11.
- 8a = 1/9, r = 1/3 ; (1/9)/(2/3) = 1/6.
Harmonic progression and the three means
If a distance is covered one way at one speed and back at another, the average speed is not the simple average; it is the harmonic mean. That is why three kinds of mean are needed. Building on classes 9–10 : Since classes 9–10 we know averages and reciprocals; an HP is just an AP of reciprocals.
A sequence whose reciprocals are in AP is a harmonic progression (HP). The harmonic mean of a and b is H = 2ab/(a + b). For positive a, b the AM = (a + b)/2 and GM = √(ab).
Rule 1: Term of an HP — The nth term of an HP = 1 ÷ (the nth term of the reciprocal AP). E.g. 1/2, 1/5, 1/8: reciprocals 2, 5, 8; the tenth reciprocal is 2 + 9 × 3 = 29; the term is 1/29.
In the figure : Semicircle: the radius is the AM = 6.5, the perpendicular is the GM = 6, and the segment CE is the HM ≈ 5.5; AM ≥ GM ≥ HM.
Rule 2: Relation of the three means — AM ≥ GM ≥ HM; AM × HM = GM²; if a = b all three are equal. From AM ≥ GM: x + 1/x ≥ 2 (x > 0), and for a fixed sum the product is greatest when the numbers are equal.
How is the formula obtained? : (√a − √b)² ≥ 0 gives a + b ≥ 2√(ab), that is AM ≥ GM. HM = 2ab/(a + b) = ab/AM and GM² = ab, so AM × HM = GM²; thus GM is the geometric mean of AM and HM, which gives HM ≤ GM.
- Easy : HM of 2 and 6 = 2 × 2 × 6/8 = 24/8 = 3.
- Medium : The 5th term of the HP 1/4, 1/7, 1/10, …: the reciprocals are 4, 7, 10, 13, 16; so 1/16.
- Exam-type : AM = 13 and GM = 12: HM = GM²/AM = 144/13. The numbers: sum 26, product 144; that is 8 and 18.
An HP has no direct formula; always take reciprocals and work as an AP.
Taking HM = (a + b)/2. The correct formula is HM = 2ab/(a + b).
A quick HM: GM² ÷ AM. Given AM and GM, the HM follows at once. Memory clue : "The middle one squared is the product of the outer two": GM² = AM × HM.
AM ≥ GM ≥ HM holds only for positive numbers; do not use it with negative numbers.
CET-style example : The least value of 9x + 4/x for x > 0? (A) 12 (B) 13 (C) 6 (D) 36
Answer : (A). AM ≥ GM: 9x + 4/x ≥ 2√(9x × 4/x) = 2 × 6 = 12. Time-saving method : When the product of the two terms is constant the least value is 2 × √product.
Spot the difference
| Mean | Formula | For 4 and 9 |
|---|---|---|
| AM | (a + b)/2 | 6.5 |
| GM | √(ab) | 6 |
| HM | 2ab/(a + b) | 72/13 |
Practice questions
- 1Find the GM of 3 and 12.
- 2Find the HM of 6 and 12.
- 3Find the 4th term of the HP 1/2, 1/4, 1/6, ….
- 4Find the HM of 9 and 25.
- 5AM = 5 and HM = 16/5; find the GM.
- 6Find the least value of x + 16/x for x > 0.
- 7If 2, b, 6 are in HP, find b.
- 8x + y = 30 with x, y positive; find the greatest value of xy.
Show answers (try them yourself first)
- 1√36 = 6.
- 22 × 72/18 = 8.
- 3व्यस्त 2, 4, 6, 8 ; पद 1/8.
- 42 × 225/34 = 225/17.
- 5GM² = 5 × 16/5 = 16 ; GM = 4.
- 6AM ≥ GM : 2√16 = 8.
- 7b = HM = 2 × 12/8 = 3.
- 8x = y = 15 ; xy = 225.
Special series and arithmetico-geometric series
If oranges are piled up, the layers can be counted as 1, 3, 6, 10 in a triangular arrangement. The totals of such arrangements come quickly from formulas. Building on classes 9–10 : Since classes 9–10 we know the sum of natural numbers; here we add the formulas for sums of squares and cubes.
The sums of the first n natural numbers: Σn, Σn², Σn³. An arithmetico-geometric series has terms of the form (AP term) × (GP term), such as 1 + 2x + 3x² + ….
Rule 1: Three formulas — Σn = n(n + 1)/2; Σn² = n(n + 1)(2n + 1)/6; Σn³ = (n(n + 1)/2)² = (Σn)².
In the figure : Dot triangles 1, 3, 6, 10; each new row has n dots.
Rule 2: Arithmetico-geometric series — For |x| < 1, 1 + 2x + 3x² + … = 1/(1 − x)², so x + 2x² + 3x³ + … = x/(1 − x)². For complicated terms split Σ into Σn, Σn², and so on.
How is the formula obtained? : S = 1 + 2x + 3x² + … and xS = x + 2x² + …. Subtracting gives S(1 − x) = 1 + x + x² + … = 1/(1 − x); so S = 1/(1 − x)². For Σn: adding the lines 1 to n and n to 1 makes every pair n + 1.
- Easy : Σn (n = 30) = 30 × 31/2 = 465.
- Medium : Σ (k² + 2k), k = 1 to 8: 204 + 2 × 36 = 276.
- Exam-type : Σn² = 285: n(n + 1)(2n + 1) = 1710 = 9 × 10 × 19, n = 9. Σn³ = (9 × 10/2)² = 45² = 2025. Also Σ n/2ⁿ = (1/2)/(1/2)² = 2.
Σn³ = (Σn)²; the sum of the first n cubes is always a perfect square.
Treating Σ(k²) as (Σk)². Σk² = 385 (n = 10) but (Σk)² = 3025.
For terms like k(k + 1) or k(k + 2), expand first and add Σk² and Σk. Memory clue : "Expand, then split".
For sums of squares of even or odd numbers take out the factor: 2² + 4² + … = 4 × Σn².
CET-style example : 1 × 3 + 2 × 4 + … + 10 × 12 = ? (A) 495 (B) 440 (C) 385 (D) 550
Answer : (A). Σk(k + 2) = Σk² + 2Σk = 385 + 2 × 55 = 495. Time-saving method : Remember Σk² = 385 and Σk = 55 for n = 10.
Spot the difference
| Sum | Formula | For n = 10 |
|---|---|---|
| Σn | n(n + 1)/2 | 55 |
| Σn² | n(n + 1)(2n + 1)/6 | 385 |
| Σn³ | (n(n + 1)/2)² | 3025 |
Practice questions
- 1Find 1 + 2 + … + 40.
- 2Find 1² + 2² + … + 6².
- 3Find 1³ + 2³ + … + 5³.
- 4Find the sum of the first 12 odd numbers.
- 5Find 2 + 4 + 6 + … + 40.
- 6Find Σ k(k + 1), k = 1 to 8.
- 7If 1³ + 2³ + … + n³ = 441, find n.
- 8Find the sum to infinity of 3/4 + 6/16 + 9/64 + ….
Show answers (try them yourself first)
- 140 × 41/2 = 820.
- 26 × 7 × 13/6 = 91.
- 3(5 × 6/2)² = 225.
- 412² = 144.
- 52 × (1 + … + 20) = 2 × 210 = 420.
- 6204 + 36 = 240.
- 7(n(n + 1)/2)² = 441 ; n(n + 1)/2 = 21 ; n = 6.
- 83 × (1/4)/(3/4)² = 3 × (1/4) × (16/9) = 4/3.
Lesson 2: Sequences and series — quick revision
| Topic | Formula / point |
|---|---|
| AP | Tₙ = a + (n − 1)d ; Sₙ = n/2 × (2a + (n − 1)d) |
| GP | Tₙ = a rⁿ⁻¹ ; Sₙ = a (rⁿ − 1)/(r − 1) |
| Infinite GP | S∞ = a/(1 − r), ∣r∣ < 1 ; recurring decimal = GP |
| HP | reciprocals in AP ; HM = 2ab/(a + b) |
| Three means | AM ≥ GM ≥ HM ; AM × HM = GM² |
| Special series | Σn = n(n + 1)/2 ; Σn² = n(n + 1)(2n + 1)/6 ; Σn³ = (Σn)² |
| Arithmetico-geometric | 1 + 2x + 3x² + … = 1/(1 − x)² |
🎯 Most important points for the exam
- Use (n − 1) in Tₙ.
- Watch the sign of r in a GP.
- Check ∣r∣ < 1 before a sum to infinity.
- For an HP always take reciprocals.
- Split the terms of Σ and use Σn, Σn².
"AP adds, GP multiplies, HP reciprocates".