Question

The sum of first n terms of an A.P. is 3n² + 4n. Find the 25^th term of this A.P.

Answer 1

We have sum of n terms, 

S_n = 3n² + 4n 

Put n = 1 

S₁ = T₁ = 3(1)² + 4 (1) = 7 

Put n = 2 

S₂ = 3(2)² + 4 (2) = 20 

T₂ = S₂ – S₁ = 20 – 7 = 13 

Put n = 3 

S₃ = 3(3)² + 4(3) = 39

T₃ = S₃ – S₂ = 39 – 20 = 19 

Therefore, first term is 7 and common difference, 

d = 13 – 7 = 6 

The 25^th term is, T_n= a + (n – 1) d 

T₂₅ = 7 + (25 – 1) x 6 

= 7 + 24 x 6 = 151