Find the sum of 3 digit natural no multiples of 11
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hello mate!...
➡️ a(first term)=110
➡️d(common difference)=11
➡️an(last term)=990
an=a+(n-1)d
990=110+(n-1)11
990-110=(n-1)11
880/11=(n-1)
80=n-1
n=81
So the no. of terms are 81...Putting all values in Sn
Sn=n/2[2a+(n-1)d]
=81/2[2×110+(81-1)(11)
=81/2[220+80×11}
=81/2[220+880)
=81/2×1100
=81×550
=44550.
hope it helps u dear.......
➡️ a(first term)=110
➡️d(common difference)=11
➡️an(last term)=990
an=a+(n-1)d
990=110+(n-1)11
990-110=(n-1)11
880/11=(n-1)
80=n-1
n=81
So the no. of terms are 81...Putting all values in Sn
Sn=n/2[2a+(n-1)d]
=81/2[2×110+(81-1)(11)
=81/2[220+80×11}
=81/2[220+880)
=81/2×1100
=81×550
=44550.
hope it helps u dear.......
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