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# If the sum of the first n terms of an AP is $$4n - n^2$$, what is the first term (that is $$S_1$$)? What is the sum of first two terms? What is the second term? Similarly find the $$3^{rd}$$, the $$10^{th}$$ and the nth terms.

Given that,
$$S_n = 4n-n^2$$
First term, a = $$S_1 = 4(1) - (1)2 = 4-1 = 3$$
Sum of first two terms = $$S_2= 4(2)-(2)2 = 8-4 = 4$$
Second term, $$a_2 = S_2 - S_1 = 4-3 = 1$$
Common difference, d = $$a_2-a = 1-3 = -2$$

Nth term, $$a_n = a+(n-1)d$$
= 3 + (n -1)(-2)
= 3 - 2n + 2
= 5 - 2n

Therefore,$$a_3 = 5 - 2(3) = 5 - 6 = -1$$
$$a_{10} = 5-2(10) = 5 - 20 = -15$$

Hence, the sum of first two terms is 4. The second term is 1.

The 3rd, the 10th, and the nth terms are -1, -15, and 5 - 2n respectively.