In a series in GP if two terms are 6 and 30.How many mimimum terms would be required between 6 and 30 so that the terms added between 6 and 30 are unique?

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william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

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'.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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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