Primes in AP

Let the twin primes be $p,p+2$ and their mean is $p+1$ so let the mean equals $T_n$ that is the $n$ -th triangular number,

$\displaystyle \begin{aligned} p+1&=\dfrac{n(n+1)}{2} \\ 2p+2&=n^2+n \\ 2p&=n^2+n-2 \\ 2p&=(n-1)(n+2)\end{aligned}$

Now L.H.S is a product of two primes ,so must be R.H.S ,

To check let $d=(n-1,n+2)$ be the gcd and so $d|\left((n+2)-(n-1)\right)$ which makes $d|3$ or $d|1$ . If the first is true then since $3$ is a prime and LHS is the product of two primes we must have $n-1=3$ since $gcd(a,b)\le\;min\{\ a,b\}$ . But this gives $p=9$ which is not a prime.

So the latter must be true that is $(n-1,n+2)=1$ . Now for that we claim both of $(n-1)$ and $(n+2)$ to be equal to LHS in some order and since $p>2$ we have $n-1=2 \implies n=3$ , $p=5$ as the only solution. So $(5,6,7)$ is the only such pair and the answer is $\boxed{5+6+7=18}$