Question

The numbers $a, b, c, d$ are in G.P. with common ratio $r$. If $\frac{1}{a^3 + b^3} + \frac{1}{b^3 + c^3} + \frac{1}{c^3 + d^3}$ are also in G.P., then the common ratio is:

• A. $r$
• B. $r^2$
• C. $r^3$
• D. $\frac{1}{r^2}$
• E. $\frac{1}{r^3}$

Hint:

For sequences involving reciprocal terms, express each term explicitly and check the ratio of consecutive terms to determine the common ratio.

Answer 1

Answer: (E)

Since $a, b, c, d$ are in a geometric progression (G.P.) with common ratio $r$, we express them as: $$b = ar, \quad c = ar^2, \quad d = ar^3$$ The given terms: $$\frac{1}{a^3 + b^3}, \quad \frac{1}{b^3 + c^3}, \quad \frac{1}{c^3 + d^3}$$ must also form a G.P.. Substituting the values in terms of $a$: $$\frac{1}{a^3 + a^3r^3}, \quad \frac{1}{a^3r^3 + a^3r^6}, \quad \frac{1}{a^3r^6 + a^3r^9}$$ Factorizing: $$\frac{1}{a^3(1 + r^3)}, \quad \frac{1}{a^3r^3(1 + r^3)}, \quad \frac{1}{a^3r^6(1 + r^3)}$$ Since the terms form a G.P., the common ratio between consecutive terms is: $$\frac{\frac{1}{a^3r^3(1 + r^3)}}{\frac{1}{a^3(1 + r^3)}} = \frac{\frac{1}{a^3r^6(1 + r^3)}}{\frac{1}{a^3r^3(1 + r^3)}}$$ which simplifies to $\frac{1}{r^3}$.