This number 25^4 has a long sequence of positive consecutive integers, the first and the last terms of which are both prime numbers. The difference between them being a divisor of 25^4. What are the first term, last term and the number of terms of this sequence? Any help would be great. Thank you.

Guest Sep 19, 2019

#1**+1 **

**Since the common difference of the sequence is 1, therefore this difference is = number of terms. Will try the common formula for arithmetic sequence, namely: N / 2 * [2*F + D*(N - 1) ] for various divisors of 25^4, which are:1 5 25 125 625 3125 15625 78125 390625 >>Total = 9 From the above divisors, we find only one of them, 625, meets the conditions in the question: 625 / 2 * [2*F + 1*(625 - 1) ], solve for F. F = 313 - This is the first term and is also a prime. 313 + 625 - 1 = 937 - This is the last term and is also a prime. [313 + 937] / 2 * 625 =25^4 - which checks out.**

Guest Sep 20, 2019