Question

If the sum of the first n term of an AP is 4n - n2 , what is the first term ? What is the sum of first two term ? What is the second term? Similarly, find the 3rd , the 10th term and the nth term

Answer 1

We have been given ,

Sn = 4n -n^2

Just Think The Given Equation Of The Sum Of " n " Terms Is correct for all the terms in the AP , So If you put n=1 You get Sum of 1 terms which is actually the value of first term a1 .

It goes like this ,

Sn = 4n - n^2

= 4(1) - (1)^2

= 4 - 1 = 3

so The First Term Is 3.

Sum Of 2 Terms Is ...

S2 = 4(2) - (2)^2 = 4

Sum of first 2 Terms means a1 + a2 ,

Now you know a1 and S2 , so S2 - a1 = a2 = 4-3= 1

Now the common difference ,

a2- a1 = d = 1-3 = -2.

Hence The Basic Method Of Doing That was this .

Solving for a10 , an , a3 we get ,

a3 = -1 ;

a10 = -15;

an = 5 - 2n ;

Answer 2

Given :-

• Sn = 4n - n²

To find :-

• Required terms = ?

Solution :-

S1 = 4n - n²

⤇ S1 = 4 × 1 - (1)²

⤇ S1 = 4 - 1

⤇ S1 = 3

S2 = 4 × 2 - (2)²

⤇ S2 = 8 - 4

⤇ S2 = 4

S3 = 4 × 3 - (3)²

⤇ S3 = 12 - 9

⤇ S3 = 3

S9 = 4 × 9 - (9)²

⤇ S9 = 36 - 81

⤇ S9 = - 45

S10 = 4 × 10 - (10)²

⤇ S10 = 40 - 100

⤇ S10 = -60

Now,

$\rm\:s_{n-1}=4(n-1)-(n-1)^2$

⤇ 4n - 4 - [(n)² - 2 × n × 1 + (1)²]

⤇ 4n - 4 - (n² - 2n + 1)

⤇ 4n - 4 - n² + 2n - 1

⤇ 4n + 2n - n² - 4 - 1

⤇ 6n - n² - 5

Then,

a2 = S2 - S1 = 4 - 3 = 1

a3 = S3 - S2 = 3 - 4 = -1

a10 = S10 - S9 = -60 - (-45) = -60 + 45 = -15

an = Sn - $\rm\:s_{n-1}$

an = (4n - n²) - (6n - n² - 5)

⤇ an = 4n - n² - 6n + n² + 5

⤇an = 4n - 6n + 5

⤇ an = -2n + 5

⤇ an = 5 - 2n

Hence the required terms wil be S1 = 3 , S2 = 4 , a2 = 1 , S3 = 3 , a3 = -1 , a10 = -15 , an = 5 - 2n.