# In a series in AP if two terms are 18 and 32.How many minimum terms a, b, c… would be required between the two to insure that a, b, c… are unique?

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### 2 Answers

In an AP, the mth and nth terms are given by:

am = a1+(m-1)d, where a1 is the first term and d is the common diffrence between the consecutive terms.

an = a1+(n-1)d.

m is assumed to be less than m.

The terms between am and an is n-m.

Therefore , to have n-m distinct terms between an and am , we should have the condition: (an-am)/d = n-m -1 , for a particular common difference d which should not be zero.

Therefore if am = 18 and an = 32 and there are r different terms between 18 and 32.

d = (an-am)/(r+1) = (32-18)/(r+1).

in particular if r = 2, then 2 terms are in beween 18 and 32. We have only 2 terms between 18 and 32, then d=(32-18)/3 = 14/3 , so that the between terms are 18+14/3 = 22 2/3. and 18+28/3 = 27 1/3.

If

An arithmetic progression is a series where the difference between consecutive terms is a constant.

In the problem given the two given terms are 18 and 32, now if we add just one term 'a' between them

32 – a = a – 18

=> 2a = 50

=> a = 25.

The rest of the terms that form the series can be written as n1 + (n-1)* 7 where n1 is the first term, 7 is the common difference and n denotes the number of the term that we are trying to find. It is not possible to insert just one term between 18 and 32 and get more than one value for 'a'.