Find Common Difference of A.P. Given G.P. & Logarithms

In summary: Hence$$2\log_ca\log_cb\ =\ \log_ca\frac{\log_ca}2\ =\ \frac12(\log_ca)^2.$$Substituting for $\log_cb$,$$4(\log_cb)^3-2(\log_cb)^2\ =\ 4\left(\frac{\log_ca}2\right)^3-2\left(\frac{\log_ca}2\right)^2\ =\ \frac12\left(\log_ca-1\right)^2\left(\log_ca+2\right).$$Finally,
  • #1
WMDhamnekar
MHB
381
28
If a,b, c, are in G.P and $\log_ba, \log_cb,\log_ac$ are in A.P. I want to find the common difference of A.P.

Answer:

After doing some computations, I stuck here. $\frac{2(\log a+\log r)}{\log a+2\log r}=\frac{2(\log a)^2+3\log r\log a +2(\log r)^2}{(\log a)^2+\log r\log a}$

How to proceed further? (a= 1st term in G.P. r= common ratio of G.P.)
Answer is $\frac32$. I don't understand how it is computed. If any member knowing further computations to arrive at the answer, may reply.
 
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  • #2
Hi Dhamnekar Winod.

I got the answer $-\dfrac5{12}$ or $0$ (the latter being the trivial case $a=b=c$). Either the given answer is wrong or (which IMHO is the more likely) you made a typo copying the question.

Assuming you copied the question correctly, however, this is my solution.

If $a$,$b$,$c$ are in GP, then

$b\ =\ \sqrt{ac}$ (assuming $a,b,c$ are all positive, which appears to be what the question is doing)​

$\implies\ \log_ab\ =\ \dfrac{1+\log_ac}2$.

If $\log_ab$, $\log_cb$, $\log_ac$ are in AP, then
$$\log_cb\ =\ \frac{\log_ab}{\log_ac}\ =\ \frac{\log_ab+\log_ac}2$$
and substituing for $\log_ab$,
$$\frac{\frac{1+\log_ac}2}{\log_ac}\ =\ \frac{\frac{1+\log_ac}2+\log_ac}2$$
which becomes
$$3\left(\log_ac\right)^2-\log_ac-2\ =\ \left(3\log_ac+2\right)\left(\log_ac-1\right)\ =\ 0$$

$\implies\ \log_ac=-\dfrac23\ \text{or}\ 1$.

In the former case, $\log_ab=\dfrac16$ and $\log_cb=-\dfrac14$ – and you can check that
$$\frac16,\ -\frac14,\ -\frac23$$
is an AP with common difference $-\dfrac5{12}$.
 
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  • #3
Olinguito said:
Hi Dhamnekar Winod.

I got the answer $-\dfrac5{12}$ or $0$ (the latter being the trivial case $a=b=c$). Either the given answer is wrong or (which IMHO is the more likely) you made a typo copying the question.

Assuming you copied the question correctly, however, this is my solution.

If $a$,$b$,$c$ are in GP, then
$b\ =\ \sqrt{ac}$ (assuming $a,b,c$ are all positive, which appears to be what the question is doing)​

$\implies\ \log_ab\ =\ \dfrac{1+\log_ac}2$.

If $\log_ab$, $\log_cb$, $\log_ac$ are in AP, then
$$\log_cb\ =\ \frac{\log_ab}{\log_ac}\ =\ \frac{\log_ab+\log_ac}2$$
and substituing for $\log_ab$,
$$\frac{\frac{1+\log_ac}2}{\log_ac}\ =\ \frac{\frac{1+\log_ac}2+\log_ac}2$$
which becomes
$$3\left(\log_ac\right)^2-\log_ac-2\ =\ \left(3\log_ac+2\right)\left(\log_ac-1\right)\ =\ 0$$

$\implies\ \log_ac=-\dfrac23\ \text{or}\ 1$.

In the former case, $\log_ab=\dfrac16$ and $\log_cb=-\dfrac14$ – and you can check that
$$\frac16,\ -\frac14,\ -\frac23$$
is an AP with common difference $-\dfrac5{12}$.


Hello,
Your work is great. But i am sorry to say that the question posted was wrong. Please read $\log_ba$ instead of $\log_ab$.
 
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  • #4
Dhamnekar Winod said:
Please read $\log_ba$ instead of $\log_ab$.
Ah, sorry it was my fault. It was I who copied your question incorrectly. (Blush)

Now $a$, $b$, $c$ form a GP $\implies$ $b^2=ac$ $\implies$
$$2\log_cb\ =\ \log_ca+1.$$
If $\log_ba$, $\log_cb$, $\log_ac$ form an AP, then
$$2\log_cb\ =\ \log_ba+\log_ac$$
(note: $x$, $y$, $z$ form an AP $\implies$ $2y=x+z$); i.e.
$$2\log_cb\ =\ \frac{\log_ca}{\log_cb}+\frac1{\log_ca}$$
converting to base $c$.

However, substituting for $\log_ca$ or $\log_cb$ (working omitted) would give cubic equations simplifying to $\log_ca=\log_cb=1$, i.e. $a=b=c$, i.e.
$$\log_ba,\log_cb,\log_ac$$
is just $1,1,1$.

So I still don’t think the common difference is $\frac32$; it’s $0$ (i.e. a constant AP).
 
  • #5
Olinguito said:
Ah, sorry it was my fault. It was I who copied your question incorrectly. (Blush)

Now $a$, $b$, $c$ form a GP $\implies$ $b^2=ac$ $\implies$
$$2\log_cb\ =\ \log_ca+1.$$
If $\log_ba$, $\log_cb$, $\log_ac$ form an AP, then
$$2\log_cb\ =\ \log_ba+\log_ac$$
(note: $x$, $y$, $z$ form an AP $\implies$ $2y=x+z$); i.e.
$$2\log_cb\ =\ \frac{\log_ca}{\log_cb}+\frac1{\log_ca}$$
converting to base $c$.

However, substituting for $\log_ca$ or $\log_cb$ (working omitted) would give cubic equations simplifying to $\log_ca=\log_cb=1$, i.e. $a=b=c$, i.e.
$$\log_ba,\log_cb,\log_ac$$
is just $1,1,1$.

So I still don’t think the common difference is $\frac32$; it’s $0$ (i.e. a constant AP).

Hello,

After some computations, I got $$4(\log_cb)^3-2(\log_cb)^2=2\log_ca\log_cb-\log_ca+\log_cb$$ How did you solve this?
 
  • #6
Dhamnekar Winod said:
I got $$4(\log_cb)^3-2(\log_cb)^2=2\log_ca\log_cb-\log_ca+\log_cb$$

You have a mixture of $\log_ca$ and $\log_cb$. Try converting everything to $\log_ca$ or to $\log_cb$, using the relation $b^2=ac$.

Since $a$, $b$, $c$ form a GP, you have square of middle term = product of the other two terms. This applies to any three consecutive terms of any GP.

Thus
$$b^2=ac\ \implies\ 2\log_cb=\log_ca+1.$$
 
  • #7
Olinguito said:
You have a mixture of $\log_ca$ and $\log_cb$. Try converting everything to $\log_ca$ or to $\log_cb$, using the relation $b^2=ac$.

Since $a$, $b$, $c$ form a GP, you have square of middle term = product of the other two terms. This applies to any three consecutive terms of any GP.

Thus
$$b^2=ac\ \implies\ 2\log_cb=\log_ca+1.$$
Hello,

Your answer is correct. $\frac32$ common difference of A.P is wrong.
 

FAQ: Find Common Difference of A.P. Given G.P. & Logarithms

What is the formula for finding the common difference of an Arithmetic Progression (A.P.) given a Geometric Progression (G.P.) and logarithms?

The formula for finding the common difference of an A.P. given a G.P. and logarithms is:
d = (logba) / (logbr) - 1
Where d is the common difference, a is the first term of the G.P., r is the common ratio of the G.P., and b is the base of the logarithm.

Can the common difference of an A.P. be negative?

Yes, the common difference of an A.P. can be negative. This means that the terms of the A.P. are decreasing instead of increasing.

How is the common difference of an A.P. related to the common ratio of the corresponding G.P.?

The common difference of an A.P. is related to the common ratio of the corresponding G.P. by the formula:
d = (logba) / (logbr) - 1
Where d is the common difference, a is the first term of the G.P., r is the common ratio of the G.P., and b is the base of the logarithm. This formula shows that the common difference is dependent on the common ratio of the G.P.

Can the common difference of an A.P. be zero?

Yes, the common difference of an A.P. can be zero. This means that all the terms of the A.P. are the same and there is no change between them.

How can I use the common difference of an A.P. to find missing terms?

To find missing terms in an A.P., you can use the formula:
an = a1 + (n-1)d
Where an is the nth term of the A.P., a1 is the first term, n is the position of the missing term, and d is the common difference. By plugging in the known values, you can solve for the missing term.

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