Question

if the sum of first n terms of an a.p is 4n^2-n,find the 12th term?

Answer 1

Sum of first n term of an A.P is 4n² - n

Sn = 4n² - n


S ( n - 1 ) = 4 ( n - 1 )² - ( n - 1 )

S ( n - 1 ) = 4 ( n² - 2n + 1 ) - n + 1

S ( n - 1 ) = 4n² - 8n + 4 - n + 1

S ( n - 1 ) = 4n² - 9n + 5

To find nth term

an = Sn - S ( n - 1 )

an = 4n² - n - ( 4n² - 9n + 5 )

an = 4n² - n - 4n² + 9n - 5

an = 8n - 5


To find 12th term

an = 8n - 5

a12 = 8 × 12 - 5

a12 = 96 - 5

a12 = 91

Answer 2

Sn = 4n² - n

Let n be 1

S1 = 4 - 1

S1 = 3

Let n = 2

S2 = 4 × 4 - 4

S2 = 16 - 4

S2 = 12

And a1 = S1

a1 = 3

a2 = S2 - S1

a2 = 12 - 3

a2 = 9

d ( common difference ) = a2 - a1

d = 9 - 3 = 6

an = a + ( n - 1 )d

a12 = 3 + 11 × 6

a12 = 3 + 66

a12 = 69