This number 105^4 has a long sequence of positive consecutive numbers summing up to 105^4. The last term of the sequence is 15,592. What is the first term and the number of terms of this sequence? Thank you for help.

Guest Jul 29, 2019

#1**+1 **

**Sum it up from 1 to 15,592 =121,563,028 121,563,028 - 105^4 =12, 403 12,403 x 2 =24,806 Sqrt(24,806) =157.499 =~158 - the first term 15,592 - 158 + 1 =15,435 - number of terms.**

Guest Jul 29, 2019

edited by
Guest
Jul 29, 2019

#2**+1 **

105^4 = 121550625

We can solve this as follows:

Suppose that the first term is 1 and the last term is 15592

Then....the sum of these terms is (15592) (15593) / 2 = 121563028

Then the difference in the sums is 121563028 - 121550625 = 12403

So....we need to find out how many of the first n positive integers sum to 12403

So we have

n ( n + 1) / 2 = 12403

n^2 + n = 24806

n^2 + n - 24806 = 0

The square root of 24806 ≈ 157....

So.... we might guess that the factoriztion of this is ( n - 157) ( n + 158) =0

Expanding this gives us n^2 + n - 24806 = 0

Taking the positive value of n shows that n = 157

So.....the first 157 positive integers sum to 12403

Which means that the sum of the integers from 158 to 15592 gives us 105^4 = 121550625

So....the first term is 158 the last is 15592 and the number of terms is 15592 - 158 + 1 = 15435

CPhill Jul 29, 2019