find the sum of 40 positive integers divisible by 6
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Answered by
2
the first 40 positive integers divisible by 6 are 6,12,18,....... upto
40 terms
the given series is in arthimetic progression with first term a=6 and common difference d=6
sum of n terms of an A.p is
n/2×{2a+(n-1)d}
: . required sum = 40/2×{2(6)+(39)6}
=20{12+234}
=20×246
=4920
40 terms
the given series is in arthimetic progression with first term a=6 and common difference d=6
sum of n terms of an A.p is
n/2×{2a+(n-1)d}
: . required sum = 40/2×{2(6)+(39)6}
=20{12+234}
=20×246
=4920
Answered by
2
40 positive integers divisible by 6 are
6,12,18.......,240
thus
s = 6+12+18+....+240
s = 6(1+2+3....+40)
sum of n natural number = n(n+1)/2
s = 6*(40)(41)/2
s = 4920
6,12,18.......,240
thus
s = 6+12+18+....+240
s = 6(1+2+3....+40)
sum of n natural number = n(n+1)/2
s = 6*(40)(41)/2
s = 4920
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