Find the sum of all natural numbers that are less than hundred and divisible by 4
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Answered by
3
sum= 4+8+12+16+......................+96
=4 (1+2+3+4+...........+24)
formula for sum of first n natural numbers is
n (n+1)/2
=4 (24)(24+1)/2
=2×24×25
=1200
=4 (1+2+3+4+...........+24)
formula for sum of first n natural numbers is
n (n+1)/2
=4 (24)(24+1)/2
=2×24×25
=1200
Answered by
0
it is an A.P.
4,8,12,16.........96,100
no. of terms=n
l=a+(n-1)d
100=4+(n+1)4
=> n=25
now,
sum of these 25 terms=s
s=n/2*(a+l)
s=25/2*(4+100)
=> s=1300
4,8,12,16.........96,100
no. of terms=n
l=a+(n-1)d
100=4+(n+1)4
=> n=25
now,
sum of these 25 terms=s
s=n/2*(a+l)
s=25/2*(4+100)
=> s=1300
Shtakshi14:
Ohhh..so sorry. the last term i.e. l = 96 b'coz no. are less than 100
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