find the sum of all natural numbers lying between 250 and 1000 which are diversible by 9
Answers
Answered by
0
Answer:
750
the correct answer
Answered by
1
Answer:
52542
Explanation:
252+261+270+..........+999
a=252,d=9,Tn=999
Tn =a+(n-1)d
999=252+(n-1)9
999 =252+9n-9
9n=999-243
9n=756
n=756/9
n=84
Sn=n/2(2a+(n-1)d
=84/2(2×252+(84-1)9)
=42(504+747)
=42×1251
=52542
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