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find the sum of all natural numbers between 250 and 1000 which are exacty divisible by 9

Asked by Mohammed Fareed...(student) , on 27/2/13

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a=252 (as it is next number divisible by 9)

d=9

l=999 (as last number divisible by 9)

to find n

999=252+(n-1)9

747=9n-9

756=9n

n=84

S84 =

84/2(252+999)= 52542

Posted by Kanishk Ameta(student)on 27/2/13

This conversation is already closed by Expert

AP : 252 , 261 , - - - - - - - - - - - - - - - - - - -  - - -  999
a = 252
d = 9
an = 999
an = a + (n-1) d
999 = 252 + (n-1) 9
999 = 252 + 9n - 9
999 = 243 +9n
999-243 = 9n
756 = 9n
756/9 = n
n = 84

Sn = n/2 (a +an)
Sn = 84/2 (252 +999)
Sn = 24 (1251)
Sn = 30024

Posted by Riya Arora(student)on 27/2/13

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