Question

find sum of all 1and 2 digit numbers divisible by 3
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Answer 1

Step-by-step explanation:

Numbers divisible by 3 (1 and 2 digit):-

3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96 and 99.

Hence, their sum= 3+6+9+12+15+18+21+24+27+30+33+36+39+42+45+48+51+54+57+60+63+66+69+72+75+78+81+84+87+90+93+96+99= 1683

Hope it helped!

Answer 2

Answer:

Here a = 3 , d=3 and l = 99

So,

a+(n-1)d = 99

3+ ( n-1 )3 = 99

3 + 3n - 3 = 99

3n = 99

n = 99/3

n = 33

So the sum of first 33 terms of the a.p.

= n/2(a+l)

= 33/2(3+99)

= 33/2(102)

= 1683