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

Apr 29, 2018

Any (a,b,c) are in arthmetic-geometric progression

#### Explanation:

Arithmetic geometric progression means that getting from one number to the next involves multiplying by a constant then adding a constant, i.e. if we are at $a$, the next value is
$m \cdot a + n$ for some given $m , n$.

This means we have formulae for $b$ and $c$:

$b = m \cdot a + n$
$c = m \cdot b + n = m \cdot \left(m \cdot a + n\right) + n = {m}^{2} a + \left(m + 1\right) n$

If we're given a specific $a$, $b$, and $c$, we can determine $m$ and $n$. We take the formula for $b$, solve for $n$ and plug that into the equation for $c$:
$n = b - m \cdot a \implies c = {m}^{2} a + \left(m + 1\right) \left(b - m \cdot a\right)$
$c = \cancel{{m}^{2} a} + m b - m a \setminus \cancel{- {m}^{2} a} + b$
$c = m b - m a + b \implies \left(c - b\right) = m \left(b - a\right) \implies m = \frac{b - a}{c - b}$

Plugging this into the equation for $n$,
$n = b - m \cdot a = b - a \cdot \frac{b - a}{c - b} = \frac{b \left(c - b\right) - a \left(b - a\right)}{c - b}$

Therefore, given ANY $a , b , c$, we get exactly find coefficients that will make them an arithmetico-geometric progression.

This can be stated in another way. There are three "degrees of freedom" for any arithmetico-geometric progression: the initial value, the multiplied constant, and the added constant. Therefore, it takes three values exactly to determine what A.G.P. is applicable.

A geometric series, on the other hand, only has two: the ratio and the initial value. This means it takes two values to see exactly what geometric sequence is and that determines everything afterwards.

Apr 30, 2018

No such condition.

#### Explanation:

In an arithmetic geometric progression, we have term-by-term multiplication of a geometric progression with the corresponding terms of an arithmetic progression, such as

$x \cdot y , \left(x + d\right) \cdot y r , \left(x + 2 d\right) \cdot y {r}^{2} , \left(x + 3 d\right) \cdot y {r}^{3} , \ldots \ldots$

and then ${n}^{t h}$ term is $\left(x + \left(n - 1\right) d\right) y {r}^{\left(n - 1\right)}$

As $x , y , r , d$ can all be different four variables

If three terms are $a , b , c$ we will have

$x \cdot y = a$; $\left(x + d\right) y r = b$ and $\left(x + 2 d\right) y {r}^{2} = c$

and given three terms and three equations,

solving for four terms is generally not possible and relation depends more on specific values of $x , y , r$ and $d$.