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Question 13. Find the 12th term from the end of the following arithmetic progressions:(ii) 3,8,13, … ,253

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  1. This is the basic and conceptual question from Chapter name- Arithmetic Progression
    Chapter number- 9
    Exercise 9.4

    In this question we have been given an arithmetic progression. And we have to find the 12th term from the end of the arithmetic progressions 3,8,13, … ,253

CBSE DHANPAT RAI publications
Class:- 10th
Understanding CBSE Mathematics
Question 13(ii)

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1 Answer

  1. Given that,

    A.P = 3,8,13, … ,253

    last term is 253

    where, a = 3 and d = (8 – 3) = 5

    an = 3 + (n – 1)5 = 253

    3 + 5n – 5 = 253

    5n = 253 + 2 = 255

    n = 255/5

    n = 51

    Hence, the A.P has 51 terms.

    therefore, the 12th term from the end is same as (51 – 12 + 1)th of the A.P which is the 40th term.

    a40 = 3 + (40 – 1)5

    = 3 + 39(5)

    = 3 + 195 = 198

    Hence, the 12th term from the end of the A.P is 198.

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