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JEE Advanced Mathematics Sequences and Series 2022 JEE Advanced 2022 (Paper 1)

JEE Advanced Mathematics Question (2022) — Solution

Question

Let a 1 , a 2 , a 3 , …  be an arithmetic progression with a 1 = 7  and common difference 8 . Let T 1 , T 2 , T 3 , …  be such that T 1 = 3  and  T n + 1 - T n = a n  fo n ≥ 1 . Then, which of the following is/are TRUE?

Options

  1. A. T 20 = 1604
  2. B. ∑ k = 1 20 T k = 10510
  3. C. T 30 = 3454
  4. D. ∑ k = 1 30 T k = 35610

Answer

C. T 30 = 3454

Step-by-step solution

Given, a n = 7 + n - 1 8  and  T 1 = 3 Also  T n + 1 = T n + a n Now  T n = T n - 1 + a n - 1 Putting  n = 1  we get,  T 2 = T 1 + a 1 Now putting  n = 2  we get, T 3 = T 2 + a 2 ⇒ T 3 = T 1 + a 1 + a 2 And so on we get, ⇒   T n + 1 = T 1 + a 1 + a 2 + … + a n ⇒   T n + 1 = T 1 + n 2 2 7 + n - 1 8 ⇒   T n + 1 = T 1 + n 4 n + 3           ⋯ 1 Now for  n = 19  we get, T 20 = 3 + 19 79 = 1504 And for  n = 29  we get, T 30 = 3 + 29 119 = 3454   option C is correct Now finding sum  we get, ∑ k = 1 20 T k = 3 + ∑ k = 2 20 T k = 3 + ∑ k = 1 19 3 + 4 n 2 + 3 n = 3 + 3 19 + 3 19 20 2 + 4 19 20 39 6 = 3 + 10507 = 10510   option B is correct And Similarly  ∑ k = 1 30 T k = 3 + ∑ k = 1 29 4 n 2 + 3 n + 3 = 35615

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