An algebra problem by Keerthi Reddy

Algebra Level 2

log ⁡ ( a 2 b ) , log ⁡ ( a 3 b 2 ) , log ⁡ ( a 4 b 3 ) , … \log\left( \dfrac {a^2}b \right), \ \log \left( \dfrac{a^3}{b^2} \right), \ \log \left(\dfrac{a^4}{b^3} \right), \ \ldots

Can the sequence above form a arithmetic progression (in that order)? If yes, what is its common difference?

No Yes, it can. The common difference is log ⁡ ( a b ) \log( ab) Yes, it can. The common difference is log ⁡ ( a b ) \log\left( \frac ab\right)

Keerthi Reddy by Keerthi Reddy , 45, India

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1 solution

Chew-Seong Cheong
May 19, 2016

The n t h n^{th} term of the sequence is t n = log ⁡ ( a n + 1 b n ) = ( n + 1 ) log ⁡ a − n log ⁡ b t_n = \log \left( \dfrac{a^{n+1}}{b^n} \right) = (n+1) \log a - n \log b , therefore, t n + 1 = ( n + 2 ) log ⁡ a − ( n + 1 ) log ⁡ b t_{n+1} = (n+2) \log a - (n+1) \log b . We note that the difference between two consequent terms is log ⁡ a − log ⁡ b = log ⁡ ( a b ) \log a - \log b = \boxed{\log \left(\dfrac{a}{b} \right)} , the common difference.

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