Question

Find the sum of all natural numbers between 300 and 500 which are divisible by 9

Answer 1

the numbers are 306,315,.......495

a=306

d=a2-a1

d=315-306

d=9

an=495

a+(n-1)d=495

306+(n-1)9=495

306+9n-9=495

297+9n=495

9n=495-297

9n=189

n=189/9

n=21

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Answer 2

The sum of all natural numbers between 300 and 500 which are divisible by 9 is given by,

The natural numbers between 300 and 500 divisible by 9 are:

306, 315, 324, ............, 495

a = 306

d = 315 - 306 = 9

an = 495

we use formula for calculating the nth term, given by,

an = a + (n - 1)d

495 = 306 + (n - 1)9

495 = 306 + 9n - 9

495 - 306 + 9 = 9n

198 = 9n

n = 22

The sum of the AP series knowing the first and last term is given by,

Sn = n/2 [ a + an ]

= 22/2 [ 306 + 495 ]

= 11 [ 801 ]

∴ Sn = 8811

Therefore, the sum of all natural numbers between 300 and 500 which are divisible by 9 is 8811