find the sum of all multiples of 5 lying between 101 and 999
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a= 105
a^n = 995
d= 5
putting the values,
995= 105 +(n-1) ×5
890= (n-1)×5
n-1 = 890/5
n-1 = 178
n =179
so, number of multiples are 179...
sum =n/2(a+l)
179/2 ×(105 +995)
179/2 ×1100
98450, is the answer:)
a^n = 995
d= 5
putting the values,
995= 105 +(n-1) ×5
890= (n-1)×5
n-1 = 890/5
n-1 = 178
n =179
so, number of multiples are 179...
sum =n/2(a+l)
179/2 ×(105 +995)
179/2 ×1100
98450, is the answer:)
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