2
An arithmetic sequence.
The following formula generalizes this pattern and can be used to find ANY term in an arithmetic sequence. a'n = a'1+ (n-1)d.
It is 58465.
To find the sum of the first 48 terms of an arithmetic sequence, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an), where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term. In this case, a1 = 2, n = 48, and an = 2 + (48-1)*2 = 96. Plugging these values into the formula, we get: S48 = 48/2 * (2 + 96) = 24 * 98 = 2352. Therefore, the sum of the first 48 terms of the given arithmetic sequence is 2352.
The difference between successive terms in an arithmetic sequence is a constant. Denote this by r. Suppose the first term is a. Then the nth term, of the sequence is given by t(n) = (a-r) + n*r or a + (n-1)*r
It is not possible to answer this question without information on whether the terms are of an arithmetic or geometric (or other) progression, and what the starting term is.
There are 5 common differences between seventh and twelfth terms, so the CD is 2.5/5 ie 0.5. First term is therefore 15 - 6 x 0.5 = 12.
An arithmetic sequence.
Multiply them together.
To find the sum of the first 20 terms of an arithmetic progression (AP), we need to first determine the common difference (d) between the terms. Given that the 6th term is 35 and the 13th term is 70, we can calculate d by subtracting the 6th term from the 13th term and dividing by the number of terms between them: (70 - 35) / (13 - 6) = 5. The formula to find the sum of the first n terms of an AP is Sn = n/2 [2a + (n-1)d], where a is the first term. Plugging in the values for a (the 1st term), d (common difference), and n (20 terms), we can calculate the sum of the first 20 terms.
The following formula generalizes this pattern and can be used to find ANY term in an arithmetic sequence. a'n = a'1+ (n-1)d.
It is 58465.
The sum of the 1st n terms is : N(3N-1)/2 Explanation : The sum from 1 to N of (3m-2) = 3 * sumFrom1toN(m) - sumFrom1toN(2) = 3 * (N*(N+1)/2) -2*N = N(3N-1)/2 For N=10 => 145
To find the sum of the first 48 terms of an arithmetic sequence, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an), where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term. In this case, a1 = 2, n = 48, and an = 2 + (48-1)*2 = 96. Plugging these values into the formula, we get: S48 = 48/2 * (2 + 96) = 24 * 98 = 2352. Therefore, the sum of the first 48 terms of the given arithmetic sequence is 2352.
The difference between successive terms in an arithmetic sequence is a constant. Denote this by r. Suppose the first term is a. Then the nth term, of the sequence is given by t(n) = (a-r) + n*r or a + (n-1)*r
Add a constant number to one term to find the next term
You can best find out how to do this by making a project. Some examples include doing the pendulum bob or making different shapes but changing the sizes.