find the sum of all natural numbers between 400 and 600 which are divisible by 11
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2
Hey
Here is your answer,
407,418,429,.........,594
a=407
d=11
An = l= 594
An= a+(n-1)d
594-407/11 = n-1
187/11=n-1
N=17+1
N=18
Sn= n/2(a+l)
=9(407+594)
=9x1001
= 9009
Here is your answer,
407,418,429,.........,594
a=407
d=11
An = l= 594
An= a+(n-1)d
594-407/11 = n-1
187/11=n-1
N=17+1
N=18
Sn= n/2(a+l)
=9(407+594)
=9x1001
= 9009
pratyanshi:
Thnqqq
sure. u deserve that.
Answered by
1
see my answer the answer is 2179089
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aw thanks
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