1. Chapter at a Glance
A bank deposit growing at 10% a year, the population of bacteria doubling every hour, the depreciation of a car, the heights of a bouncing ball — all produce lists of numbers that follow a rule. Such ordered lists are sequences, and their sums are series. You already know arithmetic progressions; this chapter revisits them briefly and concentrates on geometric progressions (GP), their sums (including infinite sums), and the arithmetic mean and geometric mean with the famous inequality AM ≥ GM.
- Sequences (finite, infinite), terms, general term; series and sigma notation.
- AP recap: aₙ and Sₙ; arithmetic mean.
- GP: common ratio, nth term, sum of n terms.
- Sum of an infinite GP when |r| < 1.
- Geometric mean; inserting GMs.
- Relation between AM and GM: A ≥ G.
- Enrichment: Σn, Σn², Σn³.
2. Sequences and Series
A sequence is a function whose domain is ℕ (or {1, 2, …, n}); a₁, a₂, a₃, … are its terms and aₙ is the general term. A sequence may be given by a formula (aₙ = n² + 1) or recursively (a₁ = a₂ = 1, aₙ = aₙ₋₁ + aₙ₋₂ — the Virahāṅka–Fibonacci sequence).
The expression a₁ + a₂ + … + aₙ is a series, written Σ (k = 1 to n) aₖ.
3. Arithmetic Progression (Recap)
| Quantity | Formula |
|---|---|
| nth term | aₙ = a + (n − 1)d |
| Sum | Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l) |
| Arithmetic mean of a, b | A = (a + b)/2 |
| n AMs between a and b | d = (b − a)/(n + 1) |
Properties: adding/subtracting a constant to each term, or multiplying each term by a constant, keeps an AP an AP.
4. Geometric Progression
A sequence is a GP if each term after the first is obtained by multiplying the previous term by a fixed non-zero number r (common ratio): a, ar, ar², ar³, …
nth term: aₙ = arⁿ⁻¹
Sum of n terms: Sₙ = a(rⁿ − 1)/(r − 1) = a(1 − rⁿ)/(1 − r) for r ≠ 1; Sₙ = na if r = 1.
Why the sum formula works: Sₙ = a + ar + … + arⁿ⁻¹ and rSₙ = ar + … + arⁿ. Subtract: (1 − r)Sₙ = a − arⁿ.
Find the 10th term and the sum of the first 10 terms of 5, 25, 125, …
Solution: a = 5, r = 5. a₁₀ = 5¹⁰ = 9765625. S₁₀ = 5(5¹⁰ − 1)/4 = 12207030.
Which term of 2, 2√2, 4, … is 128?
Solution: r = √2. 2(√2)ⁿ⁻¹ = 128 → (√2)ⁿ⁻¹ = 64 = (√2)¹² → n = 13.
₹500 is deposited at 10% compound interest per year. Find the amount after 10 years.
Solution: Amounts form a GP with r = 1.1: 500 × (1.1)¹⁰ ≈ 500 × 2.5937 ≈ ₹1296.87.
5. Sum of an Infinite GP
If |r| < 1, rⁿ → 0 as n → ∞, so
If |r| ≥ 1 the infinite sum does not exist (the terms do not shrink).
Find 1 + ½ + ¼ + ⅛ + …, and write 0.3̅ as a fraction using an infinite GP.
Solution: 1/(1 − ½) = 2. 0.333… = 0.3 + 0.03 + … = 0.3/(1 − 0.1) = 1/3.
6. Geometric Mean and AM ≥ GM
For positive a, b: G = √(ab), so that a, G, b is a GP.
To insert n GMs between a and b: r = (b/a)^(1/(n + 1)).
For positive a, b: A ≥ G, i.e. (a + b)/2 ≥ √(ab), with equality only when a = b.
Proof: A − G = (a + b)/2 − √(ab) = (√a − √b)²/2 ≥ 0.
Insert three numbers between 1 and 256 so that the five numbers form a GP.
Solution: 256 = r⁴ → r = ±4. Numbers 4, 16, 64 (or −4, 16, −64).
Find the minimum value of x + 1/x for x > 0.
Solution: AM ≥ GM: (x + 1/x)/2 ≥ √(x × 1/x) = 1 → x + 1/x ≥ 2, equality at x = 1.
The AM and GM of two positive numbers are 10 and 8. Find the numbers.
Solution: a + b = 20, ab = 64 → a, b are roots of t² − 20t + 64 = 0 → 16 and 4.
7. Enrichment: Sums of Powers
| Sum | Formula |
|---|---|
| 1 + 2 + … + n | n(n + 1)/2 |
| 1² + 2² + … + n² | n(n + 1)(2n + 1)/6 |
| 1³ + 2³ + … + n³ | [n(n + 1)/2]² |
8. Solved Problems (Board Pattern)
Find the sum to n terms of 7 + 77 + 777 + …
Solution: = (7/9)(9 + 99 + 999 + …) = (7/9)[(10 − 1) + (10² − 1) + …] = (7/9)[10(10ⁿ − 1)/9 − n] = (7/81)(10ⁿ⁺¹ − 10 − 9n).
Three numbers in GP have sum 39/10 and product 1. Find them.
Solution: Take a/r, a, ar. Product a³ = 1 → a = 1. 1/r + 1 + r = 39/10 → 10r² − 29r + 10 = 0 → r = 5/2 or 2/5. Numbers 2/5, 1, 5/2.
A ball dropped from 10 m rebounds to half its height each time. Find the total distance travelled before it comes to rest.
Solution: Down 10, then up and down 5 + 5, 2.5 + 2.5, … Total = 10 + 2(5 + 2.5 + …) = 10 + 2 × 5/(1 − ½) = 10 + 20 = 30 m.
9. Master Formula Table
| Result | Formula |
|---|---|
| AP nth term / sum | a + (n − 1)d; n/2[2a + (n − 1)d] |
| GP nth term | arⁿ⁻¹ |
| GP sum | a(rⁿ − 1)/(r − 1), r ≠ 1 |
| Infinite GP | a/(1 − r), |r| < 1 |
| AM, GM | (a + b)/2, √(ab) |
| AM ≥ GM | (a + b)/2 ≥ √(ab) |
| Numbers from A, G | roots of t² − 2At + G² = 0 |
| Three in GP | a/r, a, ar |
10. Master Summary Table
| Feature | AP | GP |
|---|---|---|
| Rule | Add d | Multiply by r |
| nth term | a + (n − 1)d | arⁿ⁻¹ |
| Sum | n/2(a + l) | a(rⁿ − 1)/(r − 1) |
| Infinite sum | Does not exist (d ≠ 0) | a/(1 − r) if |r| < 1 |
| Mean | (a + b)/2 | √(ab) |
| Three terms | a − d, a, a + d | a/r, a, ar |
11. Mind Map
- aₙ, Σ
- Recursive rules
- a + (n − 1)d
- AM
- arⁿ⁻¹, Sₙ
- S∞ = a/(1 − r)
- G = √ab
- A ≥ G
12. Tips and Tricks
- If two GP terms are given, divide them to eliminate a: aₘ/aₙ = r^(m − n).
- Recurring decimals and bouncing balls are infinite GPs.
- For "minimum of a sum with fixed product", think AM ≥ GM.
- Using S∞ = a/(1 − r) when |r| ≥ 1.
- Applying AM ≥ GM to negative numbers.
- Forgetting the ± sign when r is found from an even power.
13. Important One-liners
- In a GP, each term is the GM of its neighbours.
- An infinite GP has a finite sum only when |r| < 1.
- AM ≥ GM for positive numbers; equal only if the numbers are equal.
- x + 1/x ≥ 2 for x > 0.
- Compound interest amounts form a GP.
14. Exam-pattern Questions with Detailed Answers
The 8th term of the GP 5/2, 5/4, 5/8, … is:
- 5/256
- 5/128
- 5/512
- 5/64
Show answer
(5/2)(1/2)⁷ = 5/256.
Answer: (A)
Find the GM of 4 and 9.
Show answer
√36 = 6.
Answer: 6
Find the sum of the infinite GP 6 + 1.2 + 0.24 + …
Show answer
r = 0.2: 6/0.8 = 7.5.
Answer: 7.5
The 4th term of a GP is the square of its 2nd term, and the first term is −3. Find the 7th term.
Show answer
ar³ = (ar)² → r = a = −3. a₇ = (−3)(−3)⁶ = −2187.
Answer: −2187
How many terms of 3, 3², 3³, … are needed to give the sum 120?
Show answer
3(3ⁿ − 1)/2 = 120 → 3ⁿ = 81 → n = 4.
Answer: 4
Find the sum to n terms of 0.6 + 0.66 + 0.666 + …
Show answer
(6/9)[(1 − 0.1) + (1 − 0.01) + …] = (2/3)[n − (0.1)(1 − 0.1ⁿ)/0.9] = (2/3)[n − (1 − 10⁻ⁿ)/9].
Answer: (2/3)n − (2/27)(1 − 10⁻ⁿ)
Express 0.4̅5̅ (= 0.4545…) as a fraction using an infinite GP.
Show answer
0.45/(1 − 0.01) = 45/99 = 5/11.
Answer: 5/11
A machine costing ₹1,00,000 depreciates by 10% of its value each year. Find its value after 5 years.
Show answer
100000 × (0.9)⁵ = 100000 × 0.59049 = ₹59,049.
Answer: ₹59,049
If a, b, c, d are in GP, prove that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in GP.
Show answer
b = ar, c = ar², d = ar³. Terms: aⁿ(1 + rⁿ), aⁿrⁿ(1 + rⁿ), aⁿr²ⁿ(1 + rⁿ) — each is rⁿ times the previous. ∎
Answer: Proved (common ratio rⁿ)
If the AM and GM of the roots of a quadratic equation are 8 and 5, find the equation.
Show answer
Sum = 16, product = 25 → x² − 16x + 25 = 0.
Answer: x² − 16x + 25 = 0
The sum of the first three terms of a GP is 16 and the sum of the next three is 128. Find a, r and Sₙ.
Show answer
a(1 + r + r²) = 16, ar³(1 + r + r²) = 128 → r³ = 8 → r = 2, a = 16/7. Sₙ = (16/7)(2ⁿ − 1).
Answer: a = 16/7, r = 2, Sₙ = (16/7)(2ⁿ − 1)
Show that the ratio of the sum of the first n terms of a GP to the sum of terms from (n + 1)th to (2n)th is 1/rⁿ.
Show answer
Sₙ = a(rⁿ − 1)/(r − 1). Next n terms: S₂ₙ − Sₙ = a(r²ⁿ − rⁿ)/(r − 1) = arⁿ(rⁿ − 1)/(r − 1). Ratio = 1/rⁿ. ∎
Answer: Proved
15. Quick Revision Summary
- Sequence: ordered list; series: its sum (Σ notation).
- AP: aₙ = a + (n − 1)d; Sₙ = n/2(a + l).
- GP: aₙ = arⁿ⁻¹; Sₙ = a(rⁿ − 1)/(r − 1); S∞ = a/(1 − r) for |r| < 1.
- AM = (a + b)/2, GM = √(ab), AM ≥ GM.
- Σn = n(n + 1)/2, Σn² = n(n + 1)(2n + 1)/6, Σn³ = [n(n + 1)/2]².