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We have to find an A.P. whose $(a_1 +1)^{th} $ term is prime & $a_1\ne 1$

I mean if first term is $4$, then $5^{th}$ term should be prime

According to me it is not true for any A.P.

because $(a_1+ 1)^{th}$ term is given by $a_1\times (1 +d)$

So it can't be a prime number

Am I right with my argument? Really?

If yes , then it is such a weird thing,

If no , please provide a counter example

Please help!!!

  • 0
    for what stands $$a_1$$? – 2017-01-07
  • 0
    What are you talking about?? – 2017-01-07
  • 0
    Do you mean the $(a_1+1)^{\text{th}}$ term? What is $a$? – 2017-01-07
  • 1
    The question is not clearly stated. – 2017-01-07
  • 2
    This is hard to follow, but looks trivial. If the $(a+1)^{st}$ term is $a+aP$ where $P$ is the period of the progression, then of course it is divisible by $a$. – 2017-01-07

1 Answers 1

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If the arithmetic progression $a_1,a_2,\dots$ has a common difference of $d$ then the $n$th term is given by $a_1+(n-1)d$. Therefore the $(a_1+1)^{\text{th}}$ term is given by $a_1+a_1d=a_1(1+d)$, which is composite provided $a_1 \neq 1$ and $1+d\neq 1$ (i.e., $d \neq 0$). So excluding the case of a constant prime arithmetic progression (e.g., $5,5,5,\dots$), you are correct that the $(a_1+1)^{\text{th}}$ term cannot be prime provided $a_1\neq 1$.