Series, Progressions and the Binomial Expansion: Question 5

Worked solution

Step 1: Set up the geometric progression condition

For three consecutive terms $a$, $b$, $c$ of a geometric progression, the ratio between consecutive terms is constant, so $\dfrac{b}{a}=\dfrac{c}{b}$, which rearranges to

$b^2 = ac.$

Here $a=4$, $b=k$, $c=9$, so

$k^2 = 4\times9 = 36.$

Step 2: Solve for $k$

$k = \pm\sqrt{36} = \pm6.$

Since the question states $k>0$, we take $k=6$.

Check: with $k=6$, the three terms are $4, 6, 9$. The common ratio is $\dfrac{6}{4}=1.5$ and $\dfrac{9}{6}=1.5$, the two ratios match, confirming $4,6,9$ genuinely form a geometric progression.

Why the other options are wrong

• A ($-6$): also satisfies $k^2=36$, but is excluded by the condition $k>0$.
• C ($6.5$): this is the arithmetic mean $\frac{4+9}{2}$, which would make $4,6.5,9$ an arithmetic progression, not a geometric one.
• D ($36$): this is $k^2$, not $k$. The square root step was skipped.

Final answer

• $k = \boxed{6}$, option B.