Question

Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms.

Answer 1

To find : AP with given conditions Given that sum of n terms is 3n² 

S_n = 3n² 

Similarly, 

Sum of n - 1 terms is 3(n - 1)² 

S_{n - 1} = 3(n - 1)² 

Formula :

T_N = S_n - S_{n - 1} = 3n² - 3(n - 1)² 

Now,

Substituting n = 1 to get the first term 

a₁ = 3 

Now,

Substituting n = 2 to get the second term 

a₂ = 9 

d = a₂ - a₁ = 6 

Hence,

The series is given by a, a + d, a + 2d…which is 3, 9, 15, 21….