Any number that you choose can be the 8th term. It is easy to find a rule based on a polynomial of order 5 such that the first five numbers are as listed in the question and the chosen eighth number. There are also non-polynomial solutions. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.
Having said that, the simplest answer, based on a polynomial of order 4, is 46259.
It is 917969.
It is possible to find a polynomial of degree 5 such that it can be made to fit the pattern of the above five numbers and any number at all that is chosen to be the eighth. However, the simplest polynomial of degree 4 is Un = 36n4 - 336n3 + 1128n2 - 1568n + 741 for n = 1, 2, 3, ... and accordingly, the 8th term is 35,813.
90
The given sequence is the sequence of perfect squares starting from 1. The nth term of this sequence can be represented as n^2. Therefore, the 8th term would be 8^2, which equals 64. So, the 8th term of the sequence 1, 4, 9, 16, 25 is 64.
It is: 1 1 2 3 5 8 13 and 21 which is the 8th term
It is 917969.
77
654
The formula is 6n + 7 where n is the nth term So 8th term would be (6 x 8) + 7 = 48 + 7 = 55
If you have this series: 1,2,3,4,5,6,7,8The 8th term is 8 and the n-th term is n.But if you have this series: 2,4,6,8,10,12,14,16The 8th term is 16 and the n-th term is 2n
It is possible to find a polynomial of degree 5 such that it can be made to fit the pattern of the above five numbers and any number at all that is chosen to be the eighth. However, the simplest polynomial of degree 4 is Un = 36n4 - 336n3 + 1128n2 - 1568n + 741 for n = 1, 2, 3, ... and accordingly, the 8th term is 35,813.
90
The given sequence is the sequence of perfect squares starting from 1. The nth term of this sequence can be represented as n^2. Therefore, the 8th term would be 8^2, which equals 64. So, the 8th term of the sequence 1, 4, 9, 16, 25 is 64.
It is: 1 1 2 3 5 8 13 and 21 which is the 8th term
90
1/8th
48