What is the nth term of 104 93 82 71?

Answer 1

The nth term is given by:

t{n} = (-3x⁴ + 30x³ - 105x² + 106x + 388)/4 for n = 1, 2, 3, 4.

Alternatively, your teacher may be expecting a much simpler (also valid) solution:

Each term is obtained by subtracting 11 from the previous term:

t{n+1} = t{n} - 11

Which means that t0 = t1 + 11 = 104 + 11 = 115

→ t{n} = 115 - 11x for n = 1, 2, 3, 4

That formula is NOT valid for any other value of n

The first formula gives t1 = 104, t2 = 93, t3 = 82, t4 = 71 and gives t5 = 42, continuing with -41, -232, -603, -1244, -2263,...

The second formula gives t1 = 104, t2 = 93, t3 = 82, t4 = 71 (the same first 4 terms) but gives t5 = 60 and continues with 49, 38, 27, 16, 5, ...

Answer 2

According to Wittgenstein's Finite Rule Paradox every finite sequence of numbers can be a described in infinitely many ways and so can be continued any of these ways - some simple, some complicated but all equally valid. Conversely, it is possible to find a rule such that any number of your choice can be the nth one.For any rule that you choose, you are making some assumption about the nature of the sequence: an assumption which may not be justified. For example, you may believe that the four terms are part of an arithmetic sequence with common difference -11, and consequently conclude that the nth term is 115-11n for n = 1, 2, 3, ... . But how do you justify your assumption that the sequence was an arithmetic sequence and not based on a polynomial of fourth degree, for example?