What is the nth term of 104 93 82 71?

Answer 1

To find the nth term of the sequence 104, 93, 82, 71, we need to determine the pattern or formula that governs the sequence. In this case, the sequence is decreasing by 11 each time. Therefore, the nth term can be expressed as 115 - 11n, where n represents the position of the term in the sequence.

Answer 2

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 3

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?