Question

the sum of n terms of a sequence is 3n2+4n.find the nth term and show that sequence is an A.p

Answer 1

Sn = 3 n² + 4 n
Tn = Sn - S_n-1 
     = 3 n² + 4 n - 3 (n-1)² - 4 (n-1)
     =  6 n + 1 

d = Tn - T_n-1 =  6 n + 1 - 6(n-1) - 1 = 6
Since  d = 6 = constant, The sequence is an AP.

Answer 2

we know,
Sn = n/2{ 2a + ( n -1)d}

now, given
Sn = 3n² + 4n
=n/2{ 6n + 8}
=n/2{ 2.7 + ( n -1)6}
now you see this is same as ,above formula in which a = 7
d = 6

Tn = 7 + 6( n -1)
=6n +1

put T1 = 6×1 +1 =7
T2 = 6×2 + 1 = 13
T3 =6×3 +1 =19

13 -7 = 19 -13 =6

so, this is an AP