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.
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.
