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A geometric progression is a series where the ratio between consecutive terms is a constant.
In the problem we have the two terms given are 6 and 30. Now if we add just one term 'a' between them
30/a = a/ 6
=> a^2 = 180
=> a = 6 sqrt 5.
Also – 6 sqrt 5 is not possible as 30 > 6.
The rest of the terms that form the series can be written as
n1*r^(n-1) = n1*(6 sqrt 5)^(n-1) where n1 is the first term, sqrt 5 is the common ratio and n denotes that we are trying to find the nth term.
It is not possible to insert just one term between 6 and 30 and try to find more than one value for 'a'.
Let us take a =a1 = 6 as the starting term, an = 30 as the nth term and r the common ratio between the consecutive terms.
ar^(n-1) = 30.
If there are no terms , between 6 and 30, then the sum is unique= 6+30 =36 unique.
If there are 1 term a2 between 6 and 30, then r = x/6 = 30/x.
Therefore x = sqrt(30*6) = +6sqrt5 Or -6sqrt5.
So there are are two GPs: 6, -6sqrt5 , 30 with r = -sqrt5 and
6 , 6sqrt5 , 30 with common ratio r = +sqrt5.
Therefore the GP is not unique.
When there are 0 or even number of terms between a1 and an ,or between 6 and 30, then n is even . Then an/a1 = r^(n-1) = 30/6 = 5 has a single real solution as n-1 is odd.
So the minimum number of terms beteen 6 and 30 is zero for the sum to be unique.
Also n should be even and not odd sothat the common ratio is unique which leads to the unique sum.
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