BITSAT 2021 Paper

Sum of odd numbers between 1 and 1000 , which is divisible by
3=3+9+15+21+27+....+999=S (let)
∴ Let n be the number of terms in series and a is first term.
∴ l=a+(n−1)d,
where l is last term and d is is common difference.
999=3+(n−1)×6
n−1=

999−3

6

=

996

6

⇒ n−1=166
⇒n=167
∴ S=

n

2

[2a+(n−1)d]
=

167

2

[2×3+(167−1)×6]
=

167

2

[1002]=167×501=83667