If {gn} is a geometric sequenc with gn > 0 for all n is an element of natural number,then prove that {ln gn} is an arthnetic sequence
g = a, ar, ar^2, ar^3, ...
ln g = ln(a), ln(a) + ln(r), ln(a) + 2ln(r), ln(a) + 3ln(r), ...
so, the AS has first term = ln(a) and difference = ln(r)
ln(gn+1) - ln (gn) = ln (gn+1/gn) = ln r. The difference between each consecutive term is a constant which is ln r. Hence, {ln gn} is arithmetic.
Not clear
yes
Why did the math book go to the comedy show?
Because it wanted to find its arithmetic sequence-tic sidekick!
Now, let's prove our point.
For a geometric sequence {gn}, we know that each term is obtained by multiplying the previous term by a constant ratio, let's call it 'r'.
So, we have:
g2 = r * g1
g3 = r * g2
g4 = r * g3
...
Now, let's take the natural logarithm of both sides of these equations:
ln(g2) = ln(r * g1)
ln(g3) = ln(r * g2)
ln(g4) = ln(r * g3)
...
Using rules of logarithms, we can rewrite these equations as:
ln(g2) = ln(r) + ln(g1)
ln(g3) = ln(r) + ln(g2)
ln(g4) = ln(r) + ln(g3)
...
Now, if we subtract ln(g1) from each equation, we get:
ln(g2) - ln(g1) = ln(r)
ln(g3) - ln(g1) = ln(r)
ln(g4) - ln(g1) = ln(r)
...
And hey, what do you know? We have an arithmetic sequence! The terms {ln(gn)} are obtained by subtracting ln(g1) from each term of {ln(gn)}!
So, we've just proved that {ln(gn)} is an arithmetic sequence. Maths can be funny too, right?
To prove that {ln(gn)} is an arithmetic sequence, we need to show that the difference between consecutive terms is constant.
Let's consider the geometric sequence {gn}. We know that gn>0 for all n, which implies that ln(gn) exists for all n, assuming the natural logarithm function.
The formula for the nth term of a geometric sequence is given by:
gn = g1 * r^(n-1),
where g1 is the first term and r is the common ratio.
Now, let's take the natural logarithm of both sides of the equation:
ln(gn) = ln(g1 * r^(n-1)).
Using the properties of logarithms, we can rewrite this as:
ln(gn) = ln(g1) + ln(r^(n-1)).
Using the power rule of logarithms, we have:
ln(gn) = ln(g1) + (n-1) * ln(r).
Now, let's consider the difference between two consecutive terms in the sequence {ln(gn)}:
ln(g(n+1)) - ln(gn) = [ln(g1) + n * ln(r)] - [ln(g1) + (n-1) * ln(r)].
Simplifying this expression, we get:
ln(g(n+1)) - ln(gn) = ln(r).
Since ln(r) is a constant (dependent on the given geometric sequence), this shows that the difference between consecutive terms in the sequence {ln(gn)} is constant. Therefore, {ln(gn)} is an arithmetic sequence.