Question

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

Answer 1

The first multiple of 9 between 400 and 800 is 405 and the last multiple is 792.

We know that An = a+(n-1)*d

                         792 = 405(n-1) * 9

                         387 = (n-1) * 9

                            n = 44.

Sum = n/2 * (2a+(n-1)*d)

        = 44/2(2 * 405 + (44-1) * 9)

       = 22(810 + 387)

      = 22 * 1197

      = 26334.

Answer 2

Answer:

an=a+(n-1) d

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