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Question 1. Find the sum of the following arithmetic progressions(iii) 3, 9/2, 6, 15/2, … to 25 terms

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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. And we have to find out sum of this progression upto 25 terms.

CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 1(iii)

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1 Answer

  1. Given A.P. has first term(a) = 3,

    Common difference(d) = 9/2 – 3 = 3/2

    and number of terms(n) = 25

    So, Sum of A.P. = S25 = n[2a + (n – 1)d] / 2

     

     

    = 25[2(3) + (25 – 1)(3/2)]/2

    = 25[6 + 36]/2 = 525

    Hence, the sum of the first 25 terms of A.P. is 525.

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