Sum of numbers between 100 and 1000 , which is divisible by 9 will be

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8. Sequences and Series

easy

The sum of the numbers between $100$ and $1000$, which is divisible by $9$ will be

A

$55350$

B

$57228$

C

$97015$

D

$62140$

Solution

(a) Series $108 + 117 + …….. + 999$ is an $A.P. $

where $a = 108$, common difference $d = 9$,

$n = \frac{{999}}{9} – \frac{{99}}{9} = 111 – 11 = 100$

Hence required sum

=$\frac{{100}}{2}(108 + 999) = 50 \times 1107 = 55350$.

Standard 11

Mathematics

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