Find the sum of natural numbers from 1 to 140 which are divisible by 4
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Answered by
53
a.p is 4 , 8, 12,16.....
a= 4 D = 4 tn = 140
tn = a + (n-1) d
140 = 4+ 4n - 4
n= 35
s35= 35/2{ 2*4 +(35 - 1 ) 4}
= 35/2{8 + 136}
= 35/2{144}
=35×72
=2520
a= 4 D = 4 tn = 140
tn = a + (n-1) d
140 = 4+ 4n - 4
n= 35
s35= 35/2{ 2*4 +(35 - 1 ) 4}
= 35/2{8 + 136}
= 35/2{144}
=35×72
=2520
superiors:
it is right
yes
Answered by
11
Divisible by 4 from 1 to 140 is like;4,8,12,....140 total 35 terms;First number is 4,common difference is 4,last term is=140;Sum of all the numbers= n/2(a+T)=(35/2)(4+140)=(35/2)*144=2520
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