Question

If the sum of the first n terms of an AP is 4n − n², what is the first term (that is, S₁)? What is the sum of the first two terms? What is the second term? Similarly, find the 3^rd, the 10th, and the n^th terms.

Answer 1

Given that,

S_n = 4n − n²

First term, a = S₁ = 4(1) − (1)² 

= 4 − 1

= 3

sum of first two terms is S₂

= 4(2) − (2)²

 = 8 − 4

= 4

Second term (a₂) = S₂ - S₁ = 4 - 3 = 1

S₃ = 4(3) - (3)² = 12 - 9 = 3 

sum of first 3 terms is 3

Third term (a₃) = S₃ - S₂ = 3 - 4 = -1

S₉ = 4(9) - (9)²

= 36 - 81

= -45

S₁₀ = 4(10) - (10)²

= 40 - 100

= -60

S₁₀ - S₉ = -60 - (-45) = -15

Now, S_n = 4n - n²

Also, `S_(n - 1)` = 4(n - 1) - (n - 1)²

= 4n - 4 - [n² - 2n + 1]

= 4n - 4 - n² + 2n - 1

= 6n - n² - 5

n^th term = S_n - S_{n - 1} = (4n - n²) - (6n - n² - 5)

= 4n - n² - 6n + n² + 5

= 5 - 2n

Thus, S₁ = 3 and a₁ = 3

S₂ = 4 and a₂ = 1

S₃ = 3 and a₃ = -1

a₁₀ = -15 and a_n = 5 - 2n

Answer 2

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

S_n = 4n -n²

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

S_n = 4n -n²

S₁ = 4(1) - (1)²

= 4 - 1 

= 3

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

S₂ = 4(2) - (2)²

= 8 - 4 

= 4

Now, as we know,

an = S_n - S_n-1

So,

a₂ = S₂ - S₁

= 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,

a₃ = S₃ - S₂

= `[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`.