Question

pleasssssssssse answer

Answer 1

It says Sn(sum) = 3 n^2 
➥Sn = 3 n^2 { Sn denotes sum of n terms } 
➥S(n-1)= 3(n-1)^2 { S(n-1) denotes sum of (n-1) terms } 
➥Sn - S(n-1) = Tn = 3n^2 - (3n^2 - 6n + 3) = 6n - 3 { Tn denotes nth term } 
➥Tn = 6n-3 
Also, a = S1 = 3 
➥ 3 + (n-1)d = 6n - 3 
➥ (n-1)d = 6(n-1) 
➥ d=6 

Answer 2

Given that the sum of any number of terms is always three times the square of a number of these terms.

sn = 3n^2

Tn = s(n) - s(n-1)

     = 3n^2 - 3(n - 1)^2

     = 3n^2 - 3(n^2 + 1 - 2n)

     = 3n^2 - (3n^2 + 3 - 6n)

     = 3n^3 - 3n^3 - 3 + 6n

     = 6n - 3.


Substitute n = 1 in (1), we get

Tn = 6(1) - 3

     = 3


Substitute n = 2 in (2), we get

Tn = 6(2) - 3

     = 9


Substitute n = 3 in (1), we get

Tn = 6(3) - 3
 
     = 15.


Therefore the AP is 3,9,15...


Hope this helps!