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Sequences and Series question

2019 · 11 Jan · Shift 2 · Q33
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Sequences and Series question

2019 · 11 Jan · Shift 2 · Q33

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
If 19th term of a non-zero A.P. is zero, then its (49th term) : (29th term) is :
  1. A
    2 : 1
  2. B
    4 : 1
  3. C
    1 : 3
  4. D
    3 : 1
View written solutionFree

Correct answer: D

  1. Let the first term of the A.P. be aaa and common difference be ddd.

  2. The nnn-th term of an A.P. is Tn=a+(n−1)dT_n = a + (n-1)dTn​=a+(n−1)d

  3. Given that the 19th term is zero: T19=a+18d=0T_{19} = a + 18d = 0T19​=a+18d=0 So, a=−18da = -18da=−18d

  4. Now find the 49th term: T49=a+48d=−18d+48d=30dT_{49} = a + 48d = -18d + 48d = 30dT49​=a+48d=−18d+48d=30d

  5. Find the 29th term: T29=a+28d=−18d+28d=10dT_{29} = a + 28d = -18d + 28d = 10dT29​=a+28d=−18d+28d=10d

  6. Therefore, T49:T29=30d:10d=3:1T_{49} : T_{29} = 30d : 10d = 3 : 1T49​:T29​=30d:10d=3:1

  7. Since the A.P. is non-zero, d≠0d \neq 0d=0 here, so the ratio is well-defined.

Hence the correct option is D.

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