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Worked solution

Recover a geometric progression from two terms

The third term of a GP is 3636 and its sixth term is 2432\tfrac{243}{2}. Find its first term and common ratio.

CJC · Sequences and Series · Example 11

Answer

a=16a=16, r=32r=\tfrac32.

Compare your reasoning

Write the two known terms

u3=ar2=36(1),u6=ar5=2432(2).\begin{aligned}u_3=ar^2&=36&&\text{(1)},\\u_6=ar^5&=\frac{243}{2}&&\text{(2)}.\end{aligned}

Let a be the first term and r the common ratio. Retain the two labelled source equations.

Divide to eliminate the first term

(2)÷(1):r3=243/236=278⟹r=32\text{(2)}\div\text{(1)}:\quad r^3=\frac{243/2}{36}=\frac{27}{8}\quad\Longrightarrow\quad r=\frac32

The source callout explains that (2) divided by (1) eliminates the unknown a. Find r first, then a using either equation. The given third term is nonzero, so this division is valid.

Substitute into the third-term equation

a=36(3/2)2=16a=\frac{36}{(3/2)^2}=16

This retains the source substitution into (1).

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