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ChelseaL
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Given that the sum of the first n terms of series, s, is 9-3^2-n
(i) find the nth term of s.
Do I have to use the formula
sn = a(1-r)/1-r?
(i) find the nth term of s.
Do I have to use the formula
sn = a(1-r)/1-r?
If that's supposed to be \(\displaystyle s_n = a \left ( \frac{1 - r^n}{1-r} \right )\) then only if it's a geometric series.ChelseaL said:Do I have to use the formula
sn = a(1-r)/1-r?
The general formula for finding the nth term of a series is a_{n} = a_{1} + (n-1)d, where a_{n} is the nth term, a_{1} is the first term, and d is the common difference.
The first term can be determined by looking at the first number in the series. The common difference can be found by subtracting the first term from the second term, or any subsequent term from the previous term.
To apply the formula to this series, you will need to determine the first term and common difference. In this case, the first term is 9 and the common difference is -6 (9 - 3 = 6, 3 - 2 = 1, 2 - (-n) = 2 + n = 1). Then, plug these values into the formula a_{n} = 9 + (n-1)(-6) and simplify to get the nth term.
You can check your answer by plugging in the value of n into the formula and seeing if it matches the corresponding term in the series. For example, if you found that the 5th term is 27, you can plug in n=5 into the formula and see if you get 27 as the answer.
Yes, the formula can be used for arithmetic and geometric series, as long as the series follows a pattern and has a constant common difference or ratio.