find the sum of all natural numbers between 100 and 200 which are divisible by 4.
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According to the question,
A.P = 104,108,...............196.
= a = 104 and d = a2-a1 = 108-104 = 4
= let An = 196
An = a+(n-1)d
= 196 = 104+(n-1)4
= 196-104 = (n-1)4
= 92/4 = n-1
= n = 23+1 = 24 terms.
Sn = n/2(a+l) (l = An)
= 24/2(104+196)
= 12 . 300 = 3600.is the answer
A.P = 104,108,...............196.
= a = 104 and d = a2-a1 = 108-104 = 4
= let An = 196
An = a+(n-1)d
= 196 = 104+(n-1)4
= 196-104 = (n-1)4
= 92/4 = n-1
= n = 23+1 = 24 terms.
Sn = n/2(a+l) (l = An)
= 24/2(104+196)
= 12 . 300 = 3600.is the answer
nemo29:
first term can be 100?
100 is also divisible by 4 then why not 100??
sayyy
in this question it is said that between 100 and 200
ok
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