find the sum of the integers between 2 and 100 which are divisible by 3
if 5 arithmetic means are inserted between 7 and 25 what is the middle mean to be inserted
The multiples of 3 between 2 and 100 are
3, 6, 9, ... , 99 , an AS where a=3 and d=3
how many of those are there?
t(n) = a+(n-1)d
99 = 3 + (n-1)(3)
96 = 3n-3
99 = 3n
n = 33
so now you want the sum of those 33 arithmetic terms
S(33) = (33/2)(first + last) = (33/2)(3+99) = 1683
Your second question....
so your 7 becomes the first term, and your 25 becomes the 7th term
25 = 7 + 6d
d = 3
so your middle term would be term(4)
= a+3d = 7+9 = 16
check: 7 10 13 16 19 22 25
1584
To find the sum of the integers between 2 and 100 that are divisible by 3, we can use the concept of an arithmetic sequence along with the formula for the sum of an arithmetic series.
An arithmetic sequence is a sequence of numbers in which the difference between any two consecutive terms is constant. In this case, the sequence consists of integers that are divisible by 3.
To find the first term of the sequence, we set up the equation:
3 * n = 2
Solving for n, we find that the first term, n, is equal to 0.67. Since the first term must be an integer, we round up to 1.
Next, we need to find the last term of the sequence. We set up the equation:
3 * n = 100
Solving for n, we find that the last term, n, is equal to 33.33. Since the last term must be an integer, we round down to 33.
Therefore, the sequence of integers between 2 and 100 that are divisible by 3 is 3, 6, 9, 12, ..., 99.
To find the number of terms in this sequence, we can use the formula:
Number of terms = (Last term - First term) / Common difference + 1
Number of terms = (33 - 1) / 3 + 1 = 11
Now, we can use the formula to find the sum of an arithmetic series:
Sum = (Number of terms / 2) * (First term + Last term)
Sum = (11 / 2) * (1 + 99) = 605
Therefore, the sum of the integers between 2 and 100 that are divisible by 3 is 605.