Find the sum of all three digit natural number which are divisible by 3 and not divisible by 6
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Answered by
6
Answer:
Step-by-step explanation:
105, 111, 117.........999
a= 105
d = 6
Tn = 999
=> a + (n-1) d = 999
=> 105 + (n-1) 6 = 999
=> (n-1) 6 = 999 - 105
=> (n-1) *6 = 894
=> (n-1) = 149
=> n = 150
Sn = n[ a + Tn] /2
= 150(105 + 999) / 2
= 75 * 1104
= 82800
Sum of all three digit no. which are divisible by 3 but not by 6 is 82800
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0
Answer:
Step-by-step explanation:
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