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
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[501]
= 83667
Hence Proved.