Find the sum of all multiples of 9lying between 300 and 500?

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Solution

Here series will be = 306, 315, 324, 333,...........495

So we have the first term a = 306 and d = 9 and last term l = 495

an=a+(n-1)d

495= 306 + (n-1)9

55= 34 + (n-1)

n = 21+1

= 22

Sum = n/2(a+l)

= 22/2(306+495)

= 8811

Hope this helps!

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Find the sum of all multiples of 9lying between 300 and 500?

Here series will be = 306, 315, 324, 333,...........495 So we have the first term a = 306 and d = 9 and last term l = 495 an=a+(n-1)d 495= 306 + (n-1)9 55= 34 + (n-1) n = 21+1 = 22 Sum = n/2(a+l) = 22/2(306+495) = 8811 Hope this helps!

Here series will be = 306, 315, 324, 333,...........495

So we have the first term a = 306 and d = 9 and last term l = 495

an=a+(n-1)d

495= 306 + (n-1)9

55= 34 + (n-1)

n = 21+1

= 22

Sum = n/2(a+l)

= 22/2(306+495)

= 8811

Hope this helps!