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Question 9. If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, . . ., is 116. Find the last term.

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One of the most important and exam oriented question from Chapter name- Arithmetic Progression
Class 10th
Chapter number- 9
Exercise :- 9.6
This type of question has been asked in previous years exams.

In this question we have been given an arithmetic progression such that sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, . . ., is 116.

Now we have to find the last term.

 

CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 9

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1 Answer

  1. Given A.P. has first term(a) = 25,

    Common difference(d) = 22 – 25 = -3 and sum(Sn) = 116.

    We know sum of n terms of an A.P. is given by, Sn = n[2a + (n – 1)d] / 2.

    => 116 = n[2(25) + (n − 1)(−3)]/2

    => n[53 − 3n]/2 = 116

    => 53n – 3n2 = 232

    => 3n2 – 53n + 232 = 0

    => 3n2 – 24n – 29n + 232 = 0

    => 3n(n – 8) – 29 (n – 8) = 0

    => (3n – 29)( n – 8 ) = 0

    => n = 29/3 or n = 8

    Ignoring n = 29/3 as number of terms cannot be a fraction, so we get, n = 8.

    So, the last term is:

    a8 = a + (8 – 1)d = 25 + 7(-3) = 25 – 21 = 4

    Hence, the last term of the given A.P. is 4.

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