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Geometric Sequence

Geometric Sequence

A geometric sequence is built by repeatedly multiplying the previous term by the same number. When the common ratio r is above 1 the terms explode like a snowball, between 0 and 1 they fade toward zero, and when r is negative the signs alternate. Any term is first term × r^(n − 1), and the sum folds into one formula; remarkably, when |r| is below 1 even an endless sum settles on a single value. Here you change the first term and ratio to watch the terms grow or fade and see an infinite series converge or diverge.

Geometric Sequence — Multiply by the Same Ratio
2
2
5
💡 Geometric = sequence with constant ratio
①Adjacent terms share the same ratio (= r)
②r > 1: exponential growth
③0 < r < 1: exponential decay
④r < 0: signs alternate
nth Term & Sum
nth Term
an = a₁ · rn-1
nth term = first term × r(n-1)
Sum
Sn = a₁(rn - 1)r - 1 (r ≠ 1)
If r = 1, Sn = na₁
📐 Deriving the Sum
①S_n = a₁ + a₁r + a₁r² + ... + a₁r^{n-1}
②rS_n = a₁r + a₁r² + ... + a₁r^n
③Subtract: S_n − rS_n = a₁ − a₁r^n
④So S_n(1−r) = a₁(1−r^n)
Geometric Series — Convergence
0.5
Infinite Geometric Series
S = a₁1 - r (|r| < 1)
Converges only when |r| < 1; diverges otherwise
Geometric Mean
Geometric Mean
b² = ac (a, b, c geometric)
Square of the middle = product of the outer terms
📐 Using the Geometric Mean
①Three numbers a, b, c geometric ⟺ b² = ac
②b = ±√(ac) — watch the sign
③Equally-spaced subseries are also geometric
Wrap-up
Geometric Core
an = a₁rn-1, Sn = a₁(rn - 1)r - 1
nth term and sum in one place
🎯 Exam Points
①nth term: a_n = a₁ · r^{n-1}
②Sum: S_n = a₁(r^n − 1)/(r − 1) (r ≠ 1)
③Infinite series: |r| < 1 gives S = a₁/(1−r)
④Geometric mean: b² = ac
⑤Sign and magnitude of r govern the behavior
Worked Examples & Past Exam
Example 1
Find the 5th term of a geometric sequence with first term 3 and common ratio 2.
1
Substitute into the general term an = a₁ · rn-1.
a5 = 3 × 25-1 = 3 × 24
2
Compute.
= 3 × 16 = 48
5th term = 48
The general term multiplies the first term by the ratio (n−1) times.
Example 2
Find the sum of the first 4 terms of a geometric sequence with first term 2 and common ratio 3.
1
Substitute into the sum formula Sn = a₁(rn − 1)/(r − 1).
S4 = 2(34 - 1)3 - 1
2
Compute.
= 2(81 - 1)2 = 80
sum = 80
The sum of a geometric sequence is a₁(rn − 1)/(r − 1) for r ≠ 1.
2022 KICE mock exam Math type, adapted
Find the sum of the infinite geometric series with first term 4 and common ratio 1/2.
8
4
2
6
does not converge
① 8
1
Since |r| = 1/2 < 1 it converges; use S = a₁/(1 − r).
S = a₁1 - r = 41 - 1/2
2
Compute.
= 41/2 = 8
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Sum of Series
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