find sum of all 1and 2 digit numbers divisible by 3
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1
Answer:
1683
Step-by-step explanation:
1 and 2 digit numbers divisible by 3 -
==: 3 , 6 , 9 ........................ 99
a = 3
d = 3
an = 99
n = ??
Sn = ??
an = 99
a + (n - 1)d = 99
3 + (n - 1) 3 = 99
3 + 3n - 3 = 99
3n = 99
n = 33
Sn = n [2a + (n - 1)d]
2
Sn = 33 [2*3 + (33 - 1)3]
2
Sn = 16.5 [6 + 96]
Sn = 16.5 * 102
Sn = 1683
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