School Revise · Grade 9 · Advanced Maths (Optional) · Chapter 6
Mathematics at Advanced Level, Class 9. This optional chapter works further with the geometric progression, finds the sum of its first terms, and meets the surprising sum of a geometric series that goes on forever.
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In a geometric progression each term is the one before multiplied by a fixed number. In this chapter we recall that pattern, find a formula for the sum of the first n terms, and see how an unending progression can still add up to a finite total.
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Geometric progression a geometric progression, GP, has each term equal to the one before it times a fixed common ratio r It runs a, ar, ar², ar³, and so on, where a is the first term. The ratio of any term to the one before is always r. In real life: Money at compound interest grows as a geometric progression, each year’s amount being the last multiplied by a fixed factor. |
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Sum of the first n terms Sₙ = a(1 − rⁿ) / (1 − r), for r not equal to 1 Rather than adding every term, this formula gives the total at once from the first term a, the ratio r and the number of terms n. It is derived by subtracting r times the sum from the sum itself. In real life: It quickly totals a repayment plan or a savings scheme where each instalment grows by a fixed factor, without adding term by term. |
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Worked example 1 Question. Find the sum 2 + 4 + 8 + 16 + 32.
Answer: The sum is 62. |
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Sum to infinity if the common ratio satisfies |r| < 1, an unending GP sums to S = a / (1 − r) When each term is a fraction of the one before, the terms shrink so fast that even infinitely many of them add to a finite number. This works only when |r| is less than 1. In real life: A bouncing ball that rises to a fixed fraction of its last height travels a finite total distance, though it bounces endlessly in theory. |
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Worked example 2 Question. Find the sum of 1 + ½ + ¼ + ⅛ + … forever.
Answer: The sum is 2. |
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Worked example 3 Question. Find the sum of the first 4 terms of 3, 6, 12, 24.
Answer: The sum is 45. |
Try this chapter hands on: change the values and watch the answer update live and animate. The interactive opens right here in the lesson.
Multiplied by a fixed common ratio r.
a(1 − rⁿ)/(1 − r), where a is the first term and r the common ratio.
|r| < 1, so that the terms shrink quickly enough to add to a finite total.
2, since each term is twice the one before it.
| 1 | a = 3, r = 3, n = 4. use Sₙ = a(1 − rⁿ)/(1 − r). |
| 2 | Substitute. 3(1 − 3⁴)/(1 − 3) = 3(1 − 81)/(−2). |
| 3 | Work it out. = 3 × (−80)/(−2) = 3 × 40 = 120. |
| 1 | a = 1, r = ⅓, |r| < 1. use S = a/(1 − r). |
| 2 | Work it out. = 1/(1 − ⅓) = 1/(2/3) = 3/2. |
½, because each term is half the one before it; since |r| < 1, an unending version would have a finite sum.
| Idea | The idea |
| GP | Each term times a fixed ratio r. |
| Terms | a, ar, ar², ar³, … |
| Sum of n terms | Sₙ = a(1 − rⁿ)/(1 − r). |
| Sum to infinity | a/(1 − r) when |r| < 1. |
| Compound interest | A real GP. |
| Common ratio | Any term divided by the one before. |
| Open the Virtual Lab |
These free Grade 9 Advanced Maths notes explain geometric progressions, summing a GP, and the infinite geometric series, with clear step by step worked examples and labelled diagrams for every student using the optional Advanced Level book.
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