Q21 of 52 Page 1

Find the sum of all multiples of 9 lying between 400 and 800.

To find: 405 + 414 + 423 + … + 792


Given,


a = 405,


d = 414 – 405 = 9


an = 792


a+(n – 1)d = an


Therefore,


405+(n – 1) 9 = 792


(n – 1)9 = 792 – 405 = 387


n – 1 = 387/9 = 43


n = 43 + 1 = 44


As we have,


s44 = 44/2(405+792)


= 22×1197 = 26334


So,


Sum of all multiples of 9 lying between 400 and 800 is 26334


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