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# Question 51. Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

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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 asked to show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667 by using arithmetic progressions properties.

CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 51

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1. Odd integers between 1 and 1000 which are divisible by 3 are 3, 9, 15, . . . .999.

First term(a) = 3, common difference(d) = 9 – 3 = 6

and nth term(an) = 999.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So,

=> 999 = 3 + (n – 1)6

=> 6(n – 1) = 996

=> n – 1 = 166

=> n = 167

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So,

S167 = 167[3 + 999] / 2

= 167

= 83667

Hence Proved.

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