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Problem 3: In which of the following situations, the sequence of numbers formed will form an A.P.?(ii) The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of the remaining in the cylinder.

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This is arithmetic progression based question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise – 9.3

We have been given that the amount of air present in the cylinder when a vacuum pump removes each time 1/4 of the remaining in the cylinder and we have to find that if that sequence of numbers forms AP or not.
CBSE DHANPAT RAI PUBLICATIONS
Understanding CBSE Mathematics
Class :- 10th
Question no 3(ii)

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1 Answer

  1. Solution:

    Air removed for the first time = (1 x 1/4) = 1/4

    Remaining air = 1 – 1/4 = 3/4

    Air removed for the second time = (3/4 x 1/4) = 3/16

    Remaining air = 3/4 – 3/16 = 9/16

    Air removed for the third time = (9/16 x 1/4) = 9/64

    Remaining air = 9/16 – 9/64 = 27/64

    ∴The sequence will be 1, 3/4, 9/16, 27/64

    Here, a2 – a1 = 3/4 – (1) = -1/4

    a3 – a2 = 9/16 – (3/4) = -3/16

    Since, the successive difference of list is not same

    ∴ The given sequence is not in A.P

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