10. Find the sum of all natural numbers from 1 to 200 which are divisible by 3.
Answers
Answer:
All natural numbers between 1 and 100, which are divisible by 3 are.
3,6,9,......99
which forms an A.P
first term of this A.P is a
1
=3
second term of this A.P is a
2
=6
last term of this A.P is a
n
=99
common difference
d=a
2
−a
1
⟹d=6−3=3 .
nth term of this A.P is given by
a
n
=a
1
+(n−1)d
put a
n
=99;a
1
=3 and d=3 in above equation we get,
⟹99=3+(n−1)3
⟹3n−3+3=99
⟹3n=99
n=
3
99
=33 number of terms in this A.P
now, sum of these n=33 terms is given by
S
n
=
2
n
(a
1
+a
n
)
put values of n=33;a
1
=3;a
n
=99 we get
S
22
=
2
33
(3+99)
⟹S
22
=
2
33
×102
⟹S
22
=33×51
⟹S
22
=1683
hence the sum of all natural numbers between 1 and 100, which are divisible by 3 is S
22
=1683
Answer:
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