10,341
S = 955 This is an arithmetic sequence, and the sum of an arithmetic sequence can be calculated as: S = n/2 x (U1 + Un) U1 is the first term (in this case 91) and Un is the last term (in this case 100). n presents the total number of terms in the sequence There are 10 numbers in this sequence (91, 92, 93, 94, 95, 96, 97, 98, 99, 100) So, the sum is : S = 10/2 x (91+100) = 955
The sum of an arithmetical sequence whose nth term is U(n) = a + (n-1)*d is S(n) = 1/2*n*[2a + (n-1)d] or 1/2*n(a + l) where l is the last term in the sequence.
Arithmetic : (First term)(last term)(act of terms)/2 Geometric : (first term)(total terms)+common ratio to the power of (1+2+...+(total terms-1))
To find the sum of all even numbers from 2 through 200, we can use the formula for the sum of an arithmetic series. Since the sequence is an arithmetic sequence with a common difference of 2, we can calculate the number of terms using the formula ((last term - first term) / common difference) + 1. In this case, the first term is 2, the last term is 200, and the common difference is 2. Plugging these values into the formula gives us ((200 - 2) / 2) + 1 = 100. The sum of an arithmetic series is given by the formula n/2 * (first term + last term), so the sum of all even numbers from 2 through 200 is 100/2 * (2 + 200) = 10100.
if repeating is allowed... 36 (6x6, for the last two digits) If not, 6 (3x2, last two digits)
6
The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.
Use arithmetic sequence which is adding the same every time. Then go for multiplier sequence and last exponential.
Given an arithmetic sequence whose first term is a, last term is l and common difference is d is:The series of partial sums, Sn, is given bySn = 1/2*n*(a + l) = 1/2*n*[2a + (n-1)*d]
875
first you jump to the last plant then the first plant then the last plant
The set of odd numbers is an arithmetic sequence. Let say that the sequence has n odd numbers where the first term is a1 and the last one is n. The formula to find the sum on nth terms for an arithmetic sequence is: Sn = (n/2)(a1 + an) or Sn = (n/2)[2a1 + (n - 1)d] where d is the common difference that for odd numbers is 2. Sn = (n/2)(2a1 + 2n - 2)
connectors of sequence
S = 955 This is an arithmetic sequence, and the sum of an arithmetic sequence can be calculated as: S = n/2 x (U1 + Un) U1 is the first term (in this case 91) and Un is the last term (in this case 100). n presents the total number of terms in the sequence There are 10 numbers in this sequence (91, 92, 93, 94, 95, 96, 97, 98, 99, 100) So, the sum is : S = 10/2 x (91+100) = 955
The sum of an arithmetical sequence whose nth term is U(n) = a + (n-1)*d is S(n) = 1/2*n*[2a + (n-1)d] or 1/2*n(a + l) where l is the last term in the sequence.
It means oldest first and newest last.
chronological order