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Sum_ap = ½ × number_of_terms × (first_term + last_term) For the first 88 multiples of three: number_of_terms = 88 first_term = 1 × 3 = 3 last_term = 88 × 3 = 264 → Sum = ½ × 88 × (3 + 264) = 44 × 267 = 11748
The common multiples of 3 and 7 are 21, 42, 63 and so on.
42 is the LCM of 3, 6 and 7. All multiples of 42 are multiples of 3, 6 and 7
There are infinitely many common multiples of 3, 5 and 7, each one 105 larger than the previous one. Or to put it another way: the common multiples of 3, 5 and 7 are the multiples of their lowest common multiple which is 105. ie their common multiples are all the multiples of 105, of which there is no end - there is an infinite number of multiples of 105 (or any other number [except zero]).
They are all multiples of 6