Find the sum of all 3 digit natural numbers which are multiples of 11
Answers
Answered by
15
110, 121, 132....... 990
a = 110
d = 11
L = 990
=> a +(n-1) d = 990
=> 110 + (n-1) 11 = 990
=> (n-1) 11 = 880
=> n-1 = 80
=> n = 81
Now,
Sn = n/2 [ 2a +(n-1) d]
= 81/2 [ 2 × 110 +(81 - 1)(11)]
= 81 /2 [ 220 + 880]
= 81 × 1100 / 2
= 81 × 550
= 44550
a = 110
d = 11
L = 990
=> a +(n-1) d = 990
=> 110 + (n-1) 11 = 990
=> (n-1) 11 = 880
=> n-1 = 80
=> n = 81
Now,
Sn = n/2 [ 2a +(n-1) d]
= 81/2 [ 2 × 110 +(81 - 1)(11)]
= 81 /2 [ 220 + 880]
= 81 × 1100 / 2
= 81 × 550
= 44550
nanditareym:
Tqu
Answered by
8
110, 121, 132....... 990
a = 110
d = 11
L = 990
=> a +(n-1) d = 990
=> 110 + (n-1) 11 = 990
=> (n-1) 11 = 880
=> n-1 = 80
=> n = 81
Now,
Sn = n/2 [ 2a +(n-1) d]
= 81/2 [ 2 × 110 +(81 - 1)(11)]
= 81 /2 [ 220 + 880]
= 81 × 1100 / 2
= 81 × 550
= 44550
hope it helped ⚡
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