IIT JEE , AIEEE Forum - Mathematics , Algebra

vishal2007

Q12:- LET a1,a2,a3................. be in A.P. WITH COMMON DIFFERENCE NOT A MULTIPLE OF 3.Then maximum number of consecutive
terms so that all are primes is
a) 2 b) 3 c) 5 d) none of these

(ans:-b)

kusum

hi,
this one is almost like a logical problem.the fact that primes do not occur so periodically after the initial few natural no.we can just think among the firdt 10 to 12 prime no and get our result.

therefore the series is 3,5,7

sudeep.kumar

well..really a great question .....

the trick is that every natural no can be written in one of the three forms..

3n, 3n+1, 3n+2.... where n is a natural no.

so lets consider any series in AP, say

let any three terms be....

a, a+d, a+2d,

then, both a and d will be eithe 3n, or 3n +1 or 3n +2

if a is of the form 3n, then it wont be a prime for any n except 3.

also we are given that d is not of the form 3n.

now let a is of the form 3n + 1 ... then d would be eithe 3n+1 or 3n +2 ...

in either case, one of the three terms of the ap is bound to be divisible by three (or three itself)

similarly if a is of the form 3n + 2, then also one of the three.. a, a+d and a+2d is divisible by three...

so if follows that if we take any three terms in ap such that the common difference is not a multiple of 3, then at least one of them is a multiple of three or 3 itself.

hence the trpilet will be all prime only if one of the terms is three...some examples of such triplets are..

3,5,7 3,7,11 3,11,19 3,13,23 3,17,31 3,23,43 and so on.

(and by the same chain of reasons it follows that no four terms of an AP can be all prime, cause even if the first of the four terms is 3, at least one of the last three terms will be divisible by 3, and so wont be a prime no.)

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