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 the six multiples of 15.
The firt three multiples of 8 are 8, 16 and 24.
77, 770 and 770000 are three.
88, 176, 264
88, 176, 264
Any and all common multiples of 8 and 11 will be divisible by the least common multiple of 8 and 11, which is 88. Therefore, the first five common multiples of 8 and 11 will be the first five multiples of 88. This gives 88, 176, 264, 352, and 440.88, 176, 264, 352, 440
11, 22, 33, 44, 55, 66, 77, 88, 99, 110
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
6,12,18 are the common multiples of 3 and 6
Multiples are found by multiplying a number by successive counting numbers. 77 x 1 = 77 77 x 2 = 154 77 x 3 = 231 77 x 4 = 308 77 x 5 = 385
The common multiples of 3 and 16 are all multiples of their LCM, which is 48. Thus, their common multiples are 48, 96, 144, 192, 240, 288, etc.
11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121
To start with the 3 common multiples of 5 they are........10,15,and 25the 3 common multiples of 3 are.........15,30,and 45hope this helped=)