The first three terms of the nth term 4n-2 would be obtained by plugging in n=1, n=2, and n=3 into the expression. Therefore, the first term would be 4(1) - 2 = 2, the second term would be 4(2) - 2 = 6, and the third term would be 4(3) - 2 = 10. So, the first three terms are 2, 6, and 10 for the given nth term 4n-2.
t(n) = 4n2 - 4n + 2
5 first terms in n²+3
If the nth term is n*7 then the first 5 terms are 7, 14, 21, 28, 35.
14112027
5, 11, 17, 23, 29
the first 4 terms of the sequence which has the nth term is a sequence of numbers that that goe together eg. 8,12,16,20,24 the nth term would be 4n+4
t(n) = 4n2 - 4n + 2
5 first terms in n²+3
no clue
nth term is 8 - n. an = 8 - n, so the sequence is {7, 6, 5, 4, 3, 2,...} (this is a decreasing sequence since the successor term is smaller than the nth term). So, the sum of first six terms of the sequence is 27.
2,1,0 is th sequence of its terms
If the nth term is n*7 then the first 5 terms are 7, 14, 21, 28, 35.
There are infinitely many possible solutions to such a question. The simplest quadratic is Un = 4n2 + 9
14112027
If the nth term is 8 -2n then the 1st four terms are 6, 4, 2, 0 and -32 is the 20th term number
The first four terms are 3 9 27 81 and 729 is the 6th term.
Yes. You can have as many nth terms as you can be bothered to write down!