Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms
To find: AP with given conditions
Given that sum of n terms is 3n2
Sn = 3n2
Similarly, sum of n - 1 terms is 3(n - 1)2
Sn - 1 = 3(n - 1)2
formula TN = Sn - Sn - 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….