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Q: Who gave the formula for finding sum of the first 'n' terms in Arithmetic Progression?

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-4 is the first negative term. The progression is 24,20,16,12,8,4,0,-4,...

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It is an Arithmetic Progression with a constant difference of 11 and first term 15.

For an Arithmetic Progression, Sum = 15[a + 7d].{a = first term and d = common difference} For a Geometric Progression, Sum = a[1-r^15]/(r-1).{r = common ratio }.

The solution to the given problem can be obtained by sum formula of arithmetic progression. In arithmetic progression difference of two consecutive terms is constant. The multiples of any whole number(in sequence) form an arithmetic progression. The first multiple of 3 is 3 and the 100th multiple is 300. 3, 6, 9, 12,... 300. There are 100 terms. The sum 3 + 6 + 9 + 12 + ... + 300 can be obtained by applying by sum formula for arithmetic progression. Sum = (N/2)(First term + Last term) where N is number of terms which in this case is 100. First term = 3; Last term = 300. Sum = (100/2)(3 + 300) = 50 x 303 = 15150.

You can use one of the formulae for the sum of an arithmetic progression to calculate that.

An arithmetic sequence is usually given by a formula in which the nth term, T(n), is given in terms of the first term, a, and the common difference, d: t(n) = a + d*(n-1) where n= 1, 2, 3, etc An alternative is to define it iteratively. Thus: t1 = a tn = tn-1 + d , where n = 2, 3, 4, etc

In an arithmetic progression the difference between each term (except the first) and the one before is a constant. In a geometric progression, their ratio is a constant. That is, Arithmetic progression U(n) - U(n-1) = d, where d, the common difference, is a constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1) + d = U(1) + (n-1)*d Geometric progression U(n) / U(n-1) = r, where r, the common ratio is a non-zero constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1)*r = U(1)*r^(n-1).

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.

Svante Arrhenius (1859-1927) in 1896 was a Swedish scientist who developed what is now know as the 'greenhouse gas law':"if the quantity of carbonic acid increases in geometric progression, the augmentation of the temperature will increase nearly in arithmetic progression"

An arithmetical set is a set of natural numbers which can be defined by a formula of first-order Peano arithmetic.

You can just go ahead and add them. Or you can use the formula for an arithmetic series.

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