find the sum of all three digit number which are multiples of 11
Answers
Answered by
8
The first three digit no. which is divisible by 11 is 110
So,
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 Ans
So,
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 Ans
Similar questions
Social Sciences,
6 months ago
Biology,
6 months ago
English,
6 months ago
Geography,
1 year ago
Math,
1 year ago