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.
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.
+128407 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
