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}, the10^{th} and the n^{th} terms.

#### Solution 1

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

d = a_{2} − a = 1 − 3 = −2

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. 3^{rd}, 10^{th}, and n^{th} terms are −1, −15, and 5 − 2n respectively.

#### Solution 2

In the given problem, the sum of *n* terms of an A.P. is given by the expression,

S_{n} = 4n -n^{2}

So here, we can find the first term by substituting n = 1 ,

S_{n} = 4n -n^{2}

S_{1} = 4(1) - (1)^{2}

= 4 - 1

= 3

Similarly, the sum of first two terms can be given by,

S_{2} = 4(2) - (2)^{2}

^{ } = 8 - 4

= 4

Now, as we know,

an = S_{n} - S_{n-1}

So,

a_{2} = S_{2} - S_{1}

_{ }= 4 - 3

= 1

Now, using the same method we have to find the third, tenth and *n*^{th} term of the A.P.

So, for the third term,

a_{3} = S_{3} - S_{2}

`=[4(3)-(3)^2]-[4(2)-(2)^2]`

` = (12-9)-(8-4)`

= 3 - 4

= - 1

Also, for the tenth term,

`a_10 = A_10 - S_9`

`=[4(10)-(10)^2]-[4(9)-(9)^2]`

= (40 - 100 ) - ( 36 - 81 )

= - 60 + 45

= - 15

So, for the *n*^{th} term,

`a_n = S_n - S_(n-1)`

`=[4(n)-(n)^2]-[4(n-1)-(n-1)^2]`

`=(4n -n^2)-(4n-4-n^2 - 1 + 2n)`

`=4n - n^2 - 4n + 4 + n^2 + 1 -2n`

= 5 - 2n

Therefore, `a = 3 , S_2 = 4 , a_2 = 1 , a_3 = -1 , a_10 = -15`.