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Question: The condition for which three numbers \((a,b,c)\) are in A.G.P is?...

The condition for which three numbers (a,b,c)(a,b,c) are in A.G.P is?

Explanation

Solution

Here, we have to find the condition for which the three numbers (a,b,c)(a,b,c) are in arithmetic geometric progression (A.G.P). In Arithmetic geometric progression each term can be represented as the product of the terms of an arithmetic progression and geometric progression. In arithmetic geometric progression we multiply a constant to the previous number and then add another constant to get the next number.

Complete step-by-step answer:
Arithmetic geometric progression can be defined as getting from one number to the next involves multiplying by a constant and then adding a constant.
So, the above statement can be written as if we are at aa, then the next value is ma+nm \cdot a + n for some given m,nm,\,n.
So, we have the formula for the next number i.e., bb and cc is
b=ma+nb = m \cdot a + n
c=mb+nc = m \cdot b + n
We can put the value of bb in c=mb+nc = m \cdot b + n from b=ma+nb = m \cdot a + n. So,
c=m(ma+n)+n\Rightarrow c = m \cdot (m \cdot a + n) + n
Multiplying by mm we get,
c=m2a+mn+n\Rightarrow c = {m^2}a + mn + n
c=m2a+n(m+1)\Rightarrow c = {m^2}a + n(m + 1)
For a given three numbers (a,b,c)(a,b,c) we can determine the value of mm and nn.
We can find the value of nn from b=ma+nb = m \cdot a + n. So,
n=bma\Rightarrow n = b - m \cdot a
Substituting the value of nn in c=m2a+n(m+1)c = {m^2}a + n(m + 1). We get,
c=m2a+(bma)(m+1)\Rightarrow c = {m^2}a + (b - m \cdot a)(m + 1)
On multiplying. We get,
c=m2a+mbmam2a+b\Rightarrow c = {m^2}a + mb - ma - {m^2}a + b
Cancelling the equal terms with opposite signs. We get,
c=mbma+b\Rightarrow c = mb - ma + b
The above equation can be written as
cb=m(ba)\Rightarrow c - b = m(b - a)
On solving we get,
m=bacb\Rightarrow m = \dfrac{{b - a}}{{c - b}}
Substitute this equation in n=bman = b - m \cdot a to get the value of nn. So,
n=b(bacb)a\Rightarrow n = b - \left( {\dfrac{{b - a}}{{c - b}}} \right) \cdot a
Solving the above equation. We get,
n=b(cb)a(ba)cb\Rightarrow n = \dfrac{{b(c - b) - a(b - a)}}{{c - b}}
So, for a given three numbers (a,b,c)(a,b,c) we need to determine the exact coefficients that will make them an arithmetic geometric progression.

Note: For arithmetic geometric progression there are three degrees of freedom: they are the initial value, the constant which we are multiplying and the constant which we are adding. Therefore, it takes three values to exactly determine the arithmetic geometric progression. In geometric series, we have only two degrees of freedom: the initial value and the ratio.