Haaa, I've literally just had this question for my maths homework.
As this is an arithmetic series, there is a common difference / the difference is always the same. For example: 1,3,5,7. The difference is always 2.
To find the common difference or in your case x, you must do the 2nd term - 1st term. Or 3rd term - 2nd term.
Therefore,
20-5x = d
3x-20=d
--> Solve the equation
20-5x=3x-20
40-5x=3x
40=8x
5=x
Thus, 5x= 25
3x=15
Hope that helps (:
An arithmetic series is the sum of the terms in an arithmetic progression.
It is 58465.
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))
Suppose the first term is a, the second is a+r and the nth is a+(n-1)r. Then the sum of the first five = 5a + 10r = 85 and the sum of the first six = 6a + 15r = 123 Solving these simultaneous equations, a = 3 and r = 7 So the first four terms are: 3, 10, 17 and 24
a1=2 d=3 an=a1+(n-1)d i.e. 2,5,8,11,14,17....
An arithmetic series is the sum of the terms in an arithmetic progression.
It is 58465.
An arithmetic series is the sequence of partial sums of an arithmetic sequence. That is, if A = {a, a+d, a+2d, ..., a+(n-1)d, ... } then the terms of the arithmetic series, S(n), are the sums of the first n terms and S(n) = n/2*[2a + (n-1)d]. Arithmetic series can never converge.A geometric series is the sequence of partial sums of a geometric sequence. That is, if G = {a, ar, ar^2, ..., ar^(n-1), ... } then the terms of the geometric series, T(n), are the sums of the first n terms and T(n) = a*(1 - r^n)/(1 - r). If |r| < 1 then T(n) tends to 1/(1 - r) as n tends to infinity.
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))
Suppose the first term is a, the second is a+r and the nth is a+(n-1)r. Then the sum of the first five = 5a + 10r = 85 and the sum of the first six = 6a + 15r = 123 Solving these simultaneous equations, a = 3 and r = 7 So the first four terms are: 3, 10, 17 and 24
a1=2 d=3 an=a1+(n-1)d i.e. 2,5,8,11,14,17....
-5 19 43 67 ...This is an arithmetic sequence because each term differs from the preceding term by a common difference, 24.In order to find the sum of the first 25 terms of the series constructed from the given arithmetic sequence, we need to use the formulaSn = [2t1 + (n - 1)d] (substitute -5 for t1, 25 for n, and 24 for d)S25 = [2(-5) + (25 - 1)24]S25 = -10 + 242S25 = 566Thus, the sum of the first 25 terms of an arithmetic series is 566.
RAMANUJANRAMANUJAN
49
To find the sum of all numbers from 51 to 150, we can use the formula for the sum of an arithmetic series: (n/2)(first term + last term), where n is the number of terms. In this case, the first term is 51, the last term is 150, and the number of terms is 150 - 51 + 1 = 100. Plugging these values into the formula, we get (100/2)(51 + 150) = 50 * 201 = 10,050. Therefore, the sum of all numbers from 51 to 150 is 10,050.
The arithmetic mean is an average arrived at by adding all the terms together and then dividing by the number of terms.
The series given is an arithmetic progression consisting of 5 terms with a common difference of 5 and first term 5 → sum{n} = (n/2)(2×5 + (n - 1)×5) = n(5n + 5)/2 = 5n(n + 1)/2 As no terms have been given beyond the 5th term, and the series is not stated to be an arithmetic progression, the above formula only holds for n = 1, 2, ..., 5.