find the sum of all 3 digit numbers which are divisible by 11.
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Answered by
9
nth term =an =990
a= 110
an = a + (n -1)d
990= 110 + (n-1) 11
880/11 = n-1
81= n
Sn =n/2 { 2a + (n-1)d}
S81= 81/2 {2×110 + (81-1)11}
= 81/2 {220 +880}
=81/2 { 1100}
=81×550
=44550
a= 110
an = a + (n -1)d
990= 110 + (n-1) 11
880/11 = n-1
81= n
Sn =n/2 { 2a + (n-1)d}
S81= 81/2 {2×110 + (81-1)11}
= 81/2 {220 +880}
=81/2 { 1100}
=81×550
=44550
Answered by
0
Answer:
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
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