Find the sum of all three digit natural number which are multiple of 9
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a1=108
d=9
an=999
an=a+(n-1)d
999=108+(n-1)9
999-108=(n-1)9
891/9=n-1
99+1=n
n=100
sn=n/2[2a+(n-1)d]
=100/2[2×108+(100-1)9
=50(216+99×9)
=50(216+891)
=50×1107
sn =55350
the sum of three digit natural number =55350
d=9
an=999
an=a+(n-1)d
999=108+(n-1)9
999-108=(n-1)9
891/9=n-1
99+1=n
n=100
sn=n/2[2a+(n-1)d]
=100/2[2×108+(100-1)9
=50(216+99×9)
=50(216+891)
=50×1107
sn =55350
the sum of three digit natural number =55350
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