Find the sum of all natural numbers between 100 and 1000 which are divisible by 11
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The 1st greatest No. greater than 100 is 110.
the last which is less than 1000 is, 990.
110,121, ………990 be the number
Tn = 90
⇒ 110 + (n – 1) 11 = 990
11n – 11 = 990 – 110
11n = 880 + 11 = 891
∴ n = 801/1 = 81
Now
S81
= 8(1/2) (110 + 990)
= 81/2(550)
= 4450
so, the sum of all number will 4450
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