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# Long Sequence

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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.

Jul 29, 2019

#1
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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.

Jul 29, 2019
edited by Guest  Jul 29, 2019
#2
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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   Jul 29, 2019