# If the sum of first n terms of an AP is 4n – n2, what is the first term i.e. S1? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd, then 10th, and nth terms.

### Solution:

Given, S_{n} = 4n – n^{2}

Therefore, S_{1} = 4(1) – (1)^{2} = 4 – 1 =** 3 (Answer)**

And S_{2} = 4(2) – (2)^{2} = 8 – 4 = 4

∴ Second term = a_{2}

= S_{2} – S_{1} [∵ a_{n} = S_{n} – S_{n – 1}]

= 4 – 3

= **1 (Answer)**

Similarly, S_{3} = 4(3) – (3)^{2} = 12 – 9 = **3(Answer)**

S_{9} = 4(9) – (9)^{2} = 36 – 81 = -45

S_{10} = 4(10) – (10)^{2} = 40 – 100 = -60

Therefore,

Third term = a_{3} = S_{3} – S_{2} [∵ a_{n} = S_{n} – S_{n – 1}]

= 3 – 4

= **-1 (Answer)**

Tenth term = a_{10} = S_{10} – S_{9} [∵ a_{n} = S_{n} – S_{n – 1}]

= -60 – (-45)

= -60 + 45

= **-15 (Answer)**

nth term = a_{n} = S_{n} – S_{n – 1}

= 4n – n^{2} – {4(n – 1) – (n – 1)^{2}}

= 4n – n^{2} – {4n – 4 – (n^{2} – 2n + 1)}

= 4n – n^{2} – {4n – 4 – n^{2} + 2n – 1}

= 4n – n^{2} – (- n^{2} + 6n – 5)

= 4n – n^{2} + n^{2} – 6n + 5

= **5 – 2n** **(Answer)**