Question

consider three digit numbers which are multiples of 6​

Answer 1

Answer:

102,108,114,120,…. 996 which is an AP with first term a= 102, common difference d= 6 and the last term l= 996.

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Answer 2

The first 3 digit multiple of 6 is 102 and the last 3 digit multiple of 6 is 996.

So the three digit multiples of 6 are 102,108,114,120,….996 which is an AP with first term a= 102, common difference d= 6 and the last term l= 996.

So number of terms n = (l-a)/d + 1

= (996–102)/6 + 1 = 149+1= 150