Infinite GP
When |r| < 1 the terms shrink toward zero fast enough that an endless sum converges to a finite value: S_∞ = a/(1 − r). This only works strictly for −1 < r < 1; if |r| ≥ 1 the sum runs away and is undefined. The formula is a CAT favourite because it turns recurring decimals, bouncing-ball distances and "infinite series" word problems into one division. For example 0.7̄ (0.777...) is the GP 0.7 + 0.07 + ... with a = 0.7, r = 0.1, so the sum is 0.7/0.9 = 7/9. Two reverse uses appear often: given S_∞ and a you can find r = 1 − a/S_∞, and given S_∞ and r you can find the first term a = S_∞(1 − r). A neat extension: the sum of squares of an infinite GP is itself an infinite GP with ratio r², so its sum is a²/(1 − r²).
✅ Solved examples
✏️ Practice — try these, take hints as needed
📝 Topic test — 8 questions
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Formula Reference Sheet
Core GP formulas
| nth term | a_n = a·r^(n−1) |
|---|---|
| Sum of n terms (r ≠ 1) | S_n = a(r^n − 1)/(r − 1) |
| Sum of n terms (|r| < 1 form) | S_n = a(1 − r^n)/(1 − r) |
| Infinite sum (|r| < 1) | S_∞ = a/(1 − r) |
| Geometric mean of a and b | GM = √(ab) |
CAT power-tools
| Three terms in GP | a/r, a, ar (product = a³) |
|---|---|
| Ratio of two terms | a_m / a_n = r^(m − n) |
| n GMs between a and b | common ratio r = (b/a)^(1/(n+1)) |
| Sum × (r − 1) | S_n(r − 1) = a(r^n − 1) |
| Each term squared | a², a²r², a²r⁴, ... is a GP with ratio r² |