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For a geometric sequence, an=a1rn−1a_n = a_1 r^{n-1}. Given a1=9a_1=9 and a4=98a_4 = \dfrac{9}{8}. Then

a4=a1r3⇒98=9r3⇒r3=18⇒r=12.a_4 = a_1 r^{3} \Rightarrow \frac{9}{8} = 9 r^3 \Rightarrow r^3 = \frac{1}{8} \Rightarrow r = \frac{1}{2}.

We take the real cube root, yielding r=1/2r=1/2. Option A is correct. Check: a2=9⋅(1/2)=4.5a_2 = 9 \cdot (1/2)= 4.5; a3=4.5⋅(1/2)=2.25a_3 = 4.5 \cdot (1/2)= 2.25; a4=2.25⋅(1/2)=1.125=9/8a_4 = 2.25 \cdot (1/2)= 1.125 = 9/8, consistent. Pitfalls include mistakenly using a4=a1r4a_4 = a_1 r^4 (off by one in the exponent) or extracting a square root instead of a cube root. On GRE sequences, attend to indexing carefully: the exponent is n−1n-1 for the nn-th term. Also note that while cube roots can have multiple complex values, GRE contexts assume the principal real root unless otherwise specified.