Question

Can there be A.P. which consists of only prime numbers?

Answer 1

We know that prime numbers are those numbers which are divisible by only 1 and the number itself. So, first assume a general A.P of n terms (i.e. ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}},{{a}_{6}}............,{{a}_{n}}$), which consists of only prime number and then find $\left( {{a}_{1}}+1 \right)^{th}$ term of the A.P. and try to contradict your above assumption.

Complete step by step answer:
Let us first assume a general A.P., which consists of n terms and also assume that all of them are prime.
Let the A.P. be ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}},{{a}_{6}}............,{{a}_{n}}$ and the common difference of the A.P. be ‘d’.
Since, the $r^{th}$ term of the above A.P. will be given by:
${{T}_{r}}={{a}_{1}}+\left( r-1 \right)d$
Now, we will find $\left( {{a}_{1}}+1 \right)^{th}$ term of the above assumed A.P.
So, ${{T}_{{{a}_{n}}+1}}={{a}_{1}}+\left( {{a}_{1}}+1-1 \right)d$
$\Rightarrow {{T}_{{{a}_{n}}+1}}={{a}_{1}}+\left( {{a}_{1}}+1-1 \right)d$
$\Rightarrow {{T}_{{{a}_{n}}+1}}={{a}_{1}}+{{a}_{1}}d$
Now, take ${{a}_{1}}$ common from both of the terms:
$\Rightarrow {{T}_{{{a}_{1}}+1}}={{a}_{1}}\left( 1+d \right)$
Here, $\left( 1+d \right)$will always be greater than 2 as the common difference of the prime number is always greater than 1 and also ‘d’ can never be zero.
So, the $\left( {{a}_{1}}+1 \right)^{th}$ of the A.P. (${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}},{{a}_{6}}............,{{a}_{n}}$) is ${{a}_{1}}\left( 1+d \right)$.
So, ${{a}_{1}}\left( 1+d \right)$ must be a prime number, and we know that prime numbers are divisible by only 1 and the number itself.
But, we can see that ${{a}_{1}}\left( 1+d \right)$ is the multiple of two number (i.e. ${{a}_{1}}$ and $\left( 1+d \right)$), and also $\left( 1+d \right)$ is always greater than or equal to 1.
Hence, we can say that $\left( {{a}_{1}}+1 \right)^{th}$ term of the A.P. which consists of only prime number is divisible by ${{a}_{1}}$ and $\left( 1+d \right)$.
Hence, $\left( {{a}_{1}}+1 \right)^{th}$ is not a prime number.
So, this contradicts our above assumption.

Hence, it is never possible to have an A.P. which consists of only prime number.

Note: Students are required to note that if they assume any A.P which consists of prime numbers then the common difference of that A.P. will always be greater than 1 because prime number is always a whole number. And, to contradict your own assumption students are required to find such a term A.P. which is possible to convert into multiple of two or more numbers.