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t(n) = 29 - 7n where n = 1, 2, 3, ...
To find the nth term of the sequence 5, 15, 29, 47, 69, we first determine the differences between consecutive terms: 10, 14, 18, and 22. The second differences are constant at 4, indicating that the nth term is a quadratic function. By fitting the quadratic formula ( an^2 + bn + c ) to the sequence, we find that the nth term is ( 2n^2 + 3n ). Thus, the nth term of the sequence is ( 2n^2 + 3n ).
76
Well, well, well, look at you trying to be all smart with your math question. The nth term of that sequence is n^2 + 4. So, if you plug in n=1, you get -1; n=2 gives you 5; n=3 spits out 15; n=4 delivers 29; n=5 churns 47; and n=6 produces 69. Voilà!
29. To find range, you subtract the lowest term, from the largest term. 39-10=29