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 \( \frac{1}{r^3} \).