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Number and Algebra - Geometric sequences

Grade 12IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A geometric sequence is a sequence in which each term after the first is found by multiplying the previous term by a fixed, non-zero constant called the common ratio rr.

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The nn-th term (unu_n) of a geometric sequence can be determined if the first term u1u_1 and the common ratio rr are known.

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If ∣r∣>1|r| > 1, the terms of the sequence increase in magnitude (diverge). If ∣r∣<1|r| < 1, the terms decrease in magnitude and approach zero (converge).

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A geometric series is the sum of the terms of a geometric sequence. The sum of the first nn terms is denoted by SnS_n.

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A convergent geometric series has a sum to infinity S∞S_{\infty}, which occurs only when the absolute value of the common ratio is less than one (∣r∣<1|r| < 1).

📐Formulae

un=u1rn−1u_n = u_1 r^{n-1}

r=un+1unr = \frac{u_{n+1}}{u_n}

Sn=u1(rn−1)r−1=u1(1−rn)1−rS_n = \frac{u_1(r^n - 1)}{r - 1} = \frac{u_1(1 - r^n)}{1 - r}

S∞=u11−r, for ∣r∣<1S_{\infty} = \frac{u_1}{1 - r}, \text{ for } |r| < 1

💡Examples

Problem 1:

In a geometric sequence, the first term u1=5u_1 = 5 and the second term u2=15u_2 = 15. Find the 66-th term u6u_6.

Solution:

r=u2u1=155=3r = \frac{u_2}{u_1} = \frac{15}{5} = 3 u6=u1r6−1=5×35u_6 = u_1 r^{6-1} = 5 \times 3^5 u6=5×243=1215u_6 = 5 \times 243 = 1215

Explanation:

First, find the common ratio rr by dividing the second term by the first. Then, use the general term formula un=u1rn−1u_n = u_1 r^{n-1} with n=6n=6.

Problem 2:

Calculate the sum of the first 88 terms of the geometric sequence 2,6,18,…2, 6, 18, \dots

Solution:

u1=2,r=62=3u_1 = 2, r = \frac{6}{2} = 3 S8=2(38−1)3−1S_8 = \frac{2(3^8 - 1)}{3 - 1} S8=2(6561−1)2S_8 = \frac{2(6561 - 1)}{2} S8=6560S_8 = 6560

Explanation:

Identify the first term and common ratio. Apply the sum formula Sn=u1(rn−1)r−1S_n = \frac{u_1(r^n - 1)}{r - 1} for n=8n=8.

Problem 3:

An infinite geometric series has a first term u1=20u_1 = 20 and a common ratio r=0.6r = 0.6. Find the sum to infinity S∞S_{\infty}.

Solution:

S∞=u11−rS_{\infty} = \frac{u_1}{1 - r} S∞=201−0.6S_{\infty} = \frac{20}{1 - 0.6} S∞=200.4=50S_{\infty} = \frac{20}{0.4} = 50

Explanation:

Since ∣0.6∣<1|0.6| < 1, the series converges. Use the sum to infinity formula S∞=u11−rS_{\infty} = \frac{u_1}{1 - r}.