first you find the first term which is multiple of 7 which is 7
then you find the last term which is multiple of 7 is 1295
then we find n (number of terms) from the formula
Un=U1+(n-1)d
Un= last term,U1= first term,n= number of terms,d=the difference between a term and the term after it which in this case is 7
1295=7+(n-1)7
1295= 7+7n-7
1295=7n
185=n
now we have every value we need so we apply them in the formula
Sn=n/2(U1+Un) "this formula can be applied only in arithmetic sequence"
=185/2(7+1295)
= 120435
There are infinitely many multiples of 9 and it is not possible to add them all.
All the multiples of 5 between 1 and 51 are: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50 These add up to 275.
The only number between 3 and 45 which is a factor of 3 is 3. There are several multiples of 3 beyond this.
The multiple of 3 nearest to and below 1000 is 999. 999/3 = 333 The sum of all the multiples is 3 x 333 x 334/2 = 166833 The multiple of 5 nearest to and below 1000 is 995. 995/5 = 199 The sum of all the multiples is 5 x 199 x 200/2 = 99500
All multiples of 3 have digits that add up to a multiple of 3. There are 333 multiples of 3 between 1 and 1000.
56, 63, 70 and 77
this isn't a place to do your math homework.
69342
There are infinitely many multiples of 9 and it is not possible to add them all.
All the multiples of 5 between 1 and 51 are: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50 These add up to 275.
The sum of all multiples of 3 below 500 is the sum of 3 + 6 + ... + 498 = 41583The sum of all multiples of 5 below 500 is the sum of 5 + 10 + ... + 495 = 24750Depending upon the interpretation of "of 3 and 5", the answer is one of:The sum of all multiples of both 3 and 5 below 500 is the sum of 15 + 30 + ... + 495 = 8415The sum of all multiples of 3 below 500 and all multiples of 5 below 500 is 41583 + 24750 = 66333Since the multiples of both 3 and 5 (that is 15, 30, ...) have been counted twice - once in the multiples of 3 and once in the multiples of 5 - the sum of all multiples of 3 or 5 or both below 500 is 41583 + 24750 - 8415 = 57918I have emphasised the word below with regard to 500 since 500 is a multiple of 5, but 500 is not below (less than) itself, that is the multiples are of the numbers 1, 2, 3, ..., 499. If the question was intended to mean less than or equal to 500, add an extra 500 to the multiples of 5 above, making the sum of the multiples of 5 be 25250 and the final sums (1) 8415, (2) 66833, (3) 58418.
27
They are infinitely many and they form an increasing sequence the sum is infinite.
The multiples of 7 between 200 and 400 are: 203, 210, 217, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 294, 301, 308, 315, 322, 329, 336, 343, 350, 357, 364, 371, 378, 385, 392, 399. The sum of all the multiples of 7 between 200 and 400 is 8,729.
-3
The sum of three consecutive multiples of 6 is 666, the multiples are 216, 222 and 228.
All multiples of 12 are also multiples of 6 and they all can be written as the sum of nine numbers.