find the sum of all three digit number which leave the same remindrer 2 which divided by 5
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102, 107, 112..............................997
a = 102, last term (an) = 997, d = 5
an = a+ (n-1)d
997 =102 + (n-1)*5
997-102 = (n-1)*5
895/5 = n-1
179 = n-1
n = 180
Sn = n/2 (a+l)1
= 180/2 (102 +997)
= 90*1099
=98910
if this is helpful than tag me brainlist
a = 102, last term (an) = 997, d = 5
an = a+ (n-1)d
997 =102 + (n-1)*5
997-102 = (n-1)*5
895/5 = n-1
179 = n-1
n = 180
Sn = n/2 (a+l)1
= 180/2 (102 +997)
= 90*1099
=98910
if this is helpful than tag me brainlist
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