find the sum of all three numbers divisible by 2 or 5.
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Answer:
a
1
=100
a
n
=995
d=5
nth term of A.P. is,
a
n
=a
1
+(n−1)d
∴995=100+(n−1)5
∴5(n−1)=995−100
∴5(n−1)=895
∴n−1=179
∴n=180
Sum of n terms of A.P. is,
∑a=
2
n
[2a+(n−1)d]
∴∑a=
2
180
[(2×100)+(180−1)5]
∴∑a=90[200+(179×5)]
∴∑a=90[200+895]
∴∑a=90×1095
∴∑a=98550
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