Math · Precalculus · Worked example

Find a term and a sum of a geometric sequence

For the geometric sequence 3, 6, 12, 24, …, find the 8th term and the sum of the first 8 terms.

Find the common ratio

Each term is twice the one before, so a₁ = 3 and r = 2.

\frac{6}{3} = \frac{12}{6} = \frac{24}{12} = 2

Use aₙ = a₁rⁿ⁻¹ with n = 8.

3 \cdot 2^{7} = 384

Use Sₙ = a₁(1 − rⁿ)/(1 − r). With r > 1, changing the signs of the numerator and the denominator keeps the numbers positive.

\frac{3 (2^{8} - 1)}{2 - 1} = 765

The eight terms are 3, 6, 12, 24, 48, 96, 192 and 384.

3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 = 765

\begin{aligned} 3 + 6 + 12 + 24 + 48 \\ + 96 + 192 + 384 = 765 \end{aligned}

a₈ = 384, and the first 8 terms add to 765.

Find the 6th term and the sum of the first 6 terms of 48, 24, 12, ….

Show the answer and explanation

r = 1/2, so a₆ = 48(1/2)⁵ = 3/2, and S₆ = 48(1 − (1/2)⁶)/(1 − 1/2) = 96 · 63/64 = 94.5.

\begin{array}{l} 48 \cdot {(\frac{1}{2})}^{5} = \frac{3}{2} \\ \frac{48 (1 - {(\frac{1}{2})}^{6})}{1 - \frac{1}{2}} = \frac{189}{2} \end{array}

In Sequences & infinite series, change the ratio to 1/2: the terms shrink, and the running totals level off toward 6, the infinite sum 3/(1 − 1/2).

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