If a, b, c, d are in a geometric progression, | vedclass

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(B) Let $b = ar, c = ar^2, d = ar^3$. Then,$(a^3 + b^3)^{-1} = \frac{1}{a^3(1 + r^3)}$. $(b^3 + c^3)^{-1} = \frac{1}{a^3r^3(1 + r^3)}$. $(c^3 + d^3)^{

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Geometric Progression

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(B) Let $b = ar, c = ar^2, d = ar^3$. Then,$(a^3 + b^3)^{-1} = \frac{1}{a^3(1 + r^3)}$. $(b^3 + c^3)^{-1} = \frac{1}{a^3r^3(1 + r^3)}$. $(c^3 + d^3)^{-1} = \frac{1}{a^3r^6(1 + r^3)}$. Let $T_1 = \frac{1}{a^3(1 + r^3)}$,$T_2 = \frac{1}{a^3r^3(1 + r^3)}$,and $T_3 = \frac{1}{a^3r^6(1 + r^3)}$. We observe that $\frac{T_2}{T_1} = \frac{1}{r^3}$ and $\frac{T_3}{T_2} = \frac{1}{r^3}$. Since the ratio is constant,the terms are in a geometric progression.

If $a, b, c, d$ are in a geometric progression,then $(a^3 + b^3)^{-1}, (b^3 + c^3)^{-1}, (c^3 + d^3)^{-1}$ are in which progression?

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