NTA Abhyas JEE Main2020MathematicsSequences and SeriesPractice

The 50 t h term of the sequence 3 + 12 + 25 + 42 + … is

Options

A5145
B5148
C5142
D5195

Correct answer

B. 5148

Step-by-step solution

Since in the sequence the second difference is constant. So, T n = a n 2 + b n + c 3 = a + b + c 12 = 4 a + 2 b + c 25 = 9 a + 3 b + c ⇒ 9 = 3 a + b and 13 = 5 a + b ⇒ 4 = 2 a   ⇒ a = 2 ⇒ b = 3 ⇒ c = - 2   ⇒ T n = 2 n 2 + 3 n - 2 T 4 = 2 × 16 + 3 × 4 - 2 = 32 + 12 - 2 = 42 T 50 = 2 × 2500 + 3 × 50 - 2 = 5000 + 150 - 2 = 5148

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