Question

If sum to n terms of an A.P. is 3n² + 5n, while T_m = 164, then value of m is

• A. 25
• B. 26
• C. 27
• D. 28

Answer 1

Answer: (C)

27

Answer 2

S_n = 3n² + 5n {S₁ = T₁ = 8 ; d = 6 }

AP ® 8, 14, 20..........

T_m = 8 + (m – 1).6 = 164

m – 1 = $\frac { 156 } { 6 }$ ̃ m = 26 +1 = 27