# Verify if the sequence where loga, log(a^2/b), log(a^3/b^2), ... is an A.P.

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### 4 Answers

LET P =log a, loga^2/b , log(a^3/b^2),

If P is an arthimatical progression, then:

a2-a1 = a3-a2

=> loga^2/b - loga = log(a^3/b^2)- log (a^2/b)

We know that:

log a - log b = log a/b

==> log (a^2/b)/a = log (a^2/b)/(a^3/b^2)

==> log (a/b) = log (b/a)

From in the information given:

a1 = loga

a2 = log(a^2 / b)

a3 = log(a^3/b^2)

Just from looking at it you can tell that it is, and the equation is log([a^n]/[b^(n-1)]); however, to solve it formulaically we do determine whether the difference between a1 and a2, and a2 and a3 is the same.

i.e. does a2 - a1 = a3 - a2

log(a^2/b) - loga = log(a^3/b^2) - log(a^2 / b

Using what we know about logs:

2loga - logb - loga = 3loga - 2logb - 2loga + logb

loga - logb = loga - logb

Therefore a1 - a2 = a2 - a3

This sequence is indeed an arithmetic series.

We'll have at least 2 methods to prove that.

We'll verify if the difference between 2 consecutive terms of the sequence is the same.

We'll note the consecutive terms as t1, t2, t3, where:

t1 = log a

t2 = log(a^2/b)

t3 = log(a^3/b^2)

We'll calculate the difference between t2 and t1:

t2 - t1 = loga - log(a^2/b)

We'll use the quotient property of the logarithms:

loga - log(a^2/b) = log (a*b/a^2)

We'll eliminate like terms:

log (a*b/a^2) = log (b/a)

t2 - t1 = log (b/a)

We'll calculate the difference between t3 and t2:

t3 - t2 = log(a^3/b^2) - log(a^2/b)

We'll use the quotient property of the logarithms, once again:

t3 - t2 = log (a^3 * b/b^2 * a^2)

We'll eliminate like terms:

t3 - t2 = log (a/b)

We notice that the difference between t2 and t1, t3 and t2 and so on is the same quantity: log (a/b).

So, the difference is the common difference between 2 consecutive terms of the sequence and the sequence is an Arithmetical Progression.

If T1 , T2 and T3 are the successive terms of an arithmetic progression (A P) then ,

T2 - T1 = T3 - T2 , the common diffrence.

Here T2 - T1 = log(a^2/b) - log a = log (a^2/b)(1/a) = log(a/b)........(1), as log x-logy = log (x+y).

T3 - T2 = log(a^2/b^2) = log (a^2/b) = log {(a^3/b^2)(b/a^2 ) }= log(a/b)...........(2).

So (1) and it is established that the common difference between the successive terms is a/b = T2-T1 = T3-T2. So the given 3 terms are in AP.