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finding the sum of geometric progressions formula
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From Wikipedia
In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed nonzero number called the common ratio. For example, the sequence 2, 6, 18, 54, ... is a geometric progression with common ratio 3. Similarly 10, 5, 2.5, 1.25, ... is a geometric sequence with common ratio 1/2. The sum of the terms of a geometric progression is known as a geometric series.
Thus, the general form of a geometric sequence is
 a,\ ar,\ ar^2,\ ar^3,\ ar^4,\ \ldots
and that of a geometric series is
 a + ar + ar^2 + ar^3 + ar^4 + \cdots
where râ‰ 0 is the common ratio and a is a scale factor, equal to the sequence's start value.
Elementary properties
The nth term of a geometric sequence with initial value a and common ratio r is given by
 a_n = a\,r^{n1}.
Such a geometric sequence also follows the recursive relation
 a_n = r\,a_{n1} for every integer n\geq 1.
Generally, to check whether a given sequence is geometric, one simply checks whether successive entries in the sequence all have the same ratio.
The common ratio of a geometric series may be negative, resulting in an alternating sequence, with numbers switching from positive to negative and back. For instance
 1, −3, 9, −27, 81, −243, …
is a geometric sequence with common ratio −3.
The behaviour of a geometric sequence depends on the value of the common ratio.
If the common ratio is:
 Positive, the terms will all be the same sign as the initial term.
 Negative, the terms will alternate between positive and negative.
 Greater than 1, there will be exponential growth towards positive infinity.
 1, the progression is a constant sequence.
 Between −1 and 1 but not zero, there will be exponential decay towards zero.
 −1, the progression is an alternating sequence (see alternating series)
 Less than −1, for the absolute values there is exponential growth towards positive and negative infinity (due to the alternating sign).
Geometric sequences (with common ratio not equal to −1,1 or 0) show exponential growth or exponential decay, as opposed to the Linear growth (or decline) of an arithmetic progression such as 4, 15, 26, 37, 48, â€¦ (with common difference 11). This result was taken by T.R. Malthus as the mathematical foundation of his Principle of Population. Note that the two kinds of progression are related: exponentiating each term of an arithmetic progression yields a geometric progression, while taking the logarithm of each term in a geometric progression with a positive common ratio yields an arithmetic progression.
Geometric series
A geometric series is the sum of the numbers in a geometric progression:
 \sum_{k=0}^{n} ar^k = ar^0+ar^1+ar^2+ar^3+\cdots+ar^n. \,
We can find a simpler formula for this sum by multiplying both sides of the above equation by 1 − r, and we'll see that
 \begin{align}
(1r) \sum_{k=0}^{n} ar^k & = (1r)(ar^0 + ar^1+ar^2+ar^3+\cdots+ar^n) \\ & = ar^0 + ar^1+ar^2+ar^3+\cdots+ar^n \\ & {\color{White}{} = ar^0}  ar^1ar^2ar^3\cdotsar^n  ar^{n+1} \\ & = a  ar^{n+1} \end{align}
since all the other terms cancel. Rearranging (for râ‰ 1) gives the convenient formula for a geometric series:
 \sum_{k=0}^{n} ar^k = \frac{a(1r^{n+1})}{1r}.
If one were to begin the sum not from 0, but from a higher term, say m, then
 \sum_{k=m}^n ar^k=\frac{a(r^mr^{n+1})}{1r}.
Differentiating this formula with respect to r allows us to arrive at formulae for sums of the form
 \sum_{k=0}^n k^s r^k.
For example:
 \frac{d}{dr}\sum_{k=0}^nr^k = \sum_{k=1}^n kr^{k1}=
\frac{1r^{n+1}}{(1r)^2}\frac{(n+1)r^n}{1r}.
For a geometric series containing only even powers of r multiply by 1r^2:
 (1r^2) \sum_{k=0}^{n} ar^{2k} = aar^{2n+2}.
Then
 \sum_{k=0}^{n} ar^{2k} = \frac{a(1r^{2n+2})}{1r^2}.
For a series with only odd powers of r
 (1r^2) \sum_{k=0}^{n} ar^{2k+1} = arar^{2n+3}
and
 \sum_{k=0}^{n} ar^{2k+1} = \frac{ar(1r^{2n+2})}{1r^2}.
Infinite geometric series
An infinite geometric series is an infinite series whose successive terms have a common ratio. Such a series converges if and only if the absolute value of the common ratio is less than one (  r  < 1 ). Its value can then be computed from the finite sum formulae
 \sum_{k=0}^\infty ar^k = \lim_{n\to\infty}{\sum_{k=0}^{n} ar^k} = \lim_{n\to\infty}\frac{a(1r^{n+1})}{1r}= \lim_{n\to\infty}\frac{a}{1r}  \lim_{n\to\infty}{\frac{ar^{n+1}}{1r}}
Since:
 r^{n+1} \to 0 \mbox{ as } n \to \infty \mbox{ when } r < 1.
Then:
 \sum_{k=0}^\infty ar^k = \frac{a}{1r}  0 = \frac{a}{1r}
For a series containing only even powers of r,
 \sum_{k=0}^\infty ar^{2k} = \frac{a}{1r^2}
and for odd powers only,
 \sum_{k=0}^\infty ar^{2k+1} = \frac{ar}{1r^2}
In cases where the sum does not start at k = 0,
 \sum_{k=m}^\infty ar^k=\frac{ar^m}{1r}
The formulae given above are valid only for  r  < 1. The latter formula is valid in every Banach algebra, as long as the norm of r is less than one, and also in the field of padic numbers if  r _{p} < 1. As in the case for a finite sum, we can differentiate to calculate
In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. For instance, the sequence 3, 5, 7, 9, 11, 13, … is an arithmetic progression with common difference 2.
If the initial term of an arithmetic progression is a_1 and the common difference of successive members is d, then the nth term of the sequence is given by:
 \ a_n = a_1 + (n  1)d,
and in general
 \ a_n = a_m + (n  m)d.
A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression.
The behavior of the arithmetic progression depends on the common difference d. If the common difference is:
 Positive, the members (terms) will grow towards positive infinity.
 Negative, the members (terms) will grow towards negative infinity.
Sum
The sum of the members of a finite arithmetic progression is called an arithmetic series.
Expressing the arithmetic series in two different ways:
 S_n=a_1+(a_1+d)+(a_1+2d)+\cdots+(a_1+(n2)d)+(a_1+(n1)d)
 S_n=(a_n(n1)d)+(a_n(n2)d)+\cdots+(a_n2d)+(a_nd)+a_n.
Adding both sides of the two equations, all terms involving d cancel:
 \ 2S_n=n(a_1+a_n).
Dividing both sides by 2 produces a common form of the equation:
 S_n=\frac{n}{2}( a_1 + a_n).
An alternate form results from reinserting the substitution: a_n = a_1 + (n1)d:
 S_n=\frac{n}{2}[ 2a_1 + (n1)d].
In 499 CE Aryabhata, a prominent mathematicianastronomer from the classical age of Indian mathematics and Indian astronomy, gave this method in the Aryabhatiya(section 2.18) .
So, for example, the sum of the terms of the arithmetic progression given by a_{n} = 3 + (n1)(5) up to the 50th term is
 S_{50} = \frac{50}{2}[2(3) + (49)(5)] = 6,275.
Product
The product of the members of a finite arithmetic progression with an initial element a_{1}, common differences d, and n elements in total is determined in a closed expression
 a_1a_2\cdots a_n = d^n {\left(\frac{a_1}{d}\right)}^{\overline{n}} = d^n \frac{\Gamma \left(a_1/d + n\right) }{\Gamma \left( a_1 / d \right) },
where x^{\overline{n}} denotes the rising factorial and \Gamma denotes the Gamma function. (Note however that the formula is not valid when a_1/d is a negative integer or zero.)
This is a generalization from the fact that the product of the progression 1 \times 2 \times \cdots \times n is given by the factorial n! and that the product
 m \times (m+1) \times (m+2) \times \cdots \times (n2) \times (n1) \times n \,\!
for positive integers m and n is given by
 \frac{n!}{(m1)!}.
Taking the example from above, the product of the terms of the arithmetic progression given by a_{n} = 3 + (n1)(5) up to the 50th term is
 P_{50} = 5^{50} \cdot \frac{\Gamma \left(3/5 + 50\right) }{\Gamma \left( 3 / 5 \right) } \approx 3.78438 \times 10^{98}
Consider an AP
a,(a+d),(a+2d),.................(a+(n1)d)
Finding the product of first three terms
a(a+d)(a+2d) =(a^{2}+ad)(a+2d) =a^{3}+3a^{2}d+2ad^{2}
this is of the form
a^{n} + na^{n1} d^{n2} + (n1)a^{n2}d^{n1}
so the product of n terms of an AP is:
a^{n} + na^{n1} d^{n2} + (n1)a^{n2}d^{n1} no solutions
From Yahoo Answers
Answers:Sn = a1(1r^n)/(1r) Sn is the sum of the first n terms in a sequence a1 is the first term in the sequence r is the common ratio in the geometric sequence n is the number of terms you are adding up so n=4 S4 = 3*(13^4)/(13) S4 = 3*80/2 S4 = 120
Answers:Sn = a + ar + ar ar^(n  1) r Sn = __ar + ar + ar^n ( 1  r ) Sn = a(1  r^n) Sn = a (1  r^n) / (1  r)
Answers:Sn = n/2[2a + (n 1)d] Sn is sum of n terms n = total numbers a = 1st term d = common difference 
Answers:In terms of the common ratio between the terms, the sum is: S = 10/(1  r). From here, the minimum possible value occurs at r = 0 with: S = 10/(1  0) = 10. I hope this helps!
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