if the answer, for example, was "FIND THE NTH TERM OF THESE NUMBERS" and they listed:
1, 1/4, 1/9, 1/16, 1/25...
then this is how I personally would go about it.
okay, so the first term is 1. write them all down like this:
1. 1
2. 1/4
3. 1/9
4. 1/16
5. 1/25
(it makes it easier to see which term is which)
as they are all over ONE, you can ignore the top number and focus on the bottom one --> and remember that 1 = 1/1.
square the N number; what do you get?
1*1 = 1
2*2 = 4
3*3 = 9
4*4 = 16
5*5 = 25
so we now have n^2 (N squared), but they aren't a fraction. *sad face*
this is where we take the expression N squared and we put a 1 over it (since all of the terms are over 1:
1/n^2 (one over N squared)
if the answers were all over 2, put -- 2 over [the expression to solve the denominator (bottom number)] -- and so on.
hope this helped :)
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my nth term maths is very tuff because its syallabus is changed
You don't.
The nth term is 7n-3 and so the next term will be 39
Oh, dude, it's like a pattern party! So, to find the nth term for this sequence, you first need to figure out the pattern. Looks like each number is decreasing by 2. So, the nth term would be 13 - 2n. Easy peasy, right?
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 ).