finding the sum of geometric progressions formula

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From Wikipedia

Geometric progression

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 non-zero 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 r≠ 0 is the common ratio and a is a scale factor, equal to the sequence's start value.

Elementary properties

The n-th term of a geometric sequence with initial value a and common ratio r is given by

a_n = a\,r^{n-1}.

Such a geometric sequence also follows the recursive relation

a_n = r\,a_{n-1} 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, … (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


(1-r) \sum_{k=0}^{n} ar^k & = (1-r)(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^1-ar^2-ar^3-\cdots-ar^n - ar^{n+1} \\ & = a - ar^{n+1} \end{align}

since all the other terms cancel. Rearranging (for r≠ 1) gives the convenient formula for a geometric series:

\sum_{k=0}^{n} ar^k = \frac{a(1-r^{n+1})}{1-r}.

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^m-r^{n+1})}{1-r}.

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^{k-1}=


For a geometric series containing only even powers of r multiply by 1-r^2:

(1-r^2) \sum_{k=0}^{n} ar^{2k} = a-ar^{2n+2}.


\sum_{k=0}^{n} ar^{2k} = \frac{a(1-r^{2n+2})}{1-r^2}.

For a series with only odd powers of r

(1-r^2) \sum_{k=0}^{n} ar^{2k+1} = ar-ar^{2n+3}


\sum_{k=0}^{n} ar^{2k+1} = \frac{ar(1-r^{2n+2})}{1-r^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 (&nbsp;|&nbsp;r&nbsp;|&nbsp;<&nbsp;1&nbsp;). 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(1-r^{n+1})}{1-r}= \lim_{n\to\infty}\frac{a}{1-r} - \lim_{n\to\infty}{\frac{ar^{n+1}}{1-r}}


r^{n+1} \to 0 \mbox{ as } n \to \infty \mbox{ when } |r| < 1.


\sum_{k=0}^\infty ar^k = \frac{a}{1-r} - 0 = \frac{a}{1-r}

For a series containing only even powers of r,

\sum_{k=0}^\infty ar^{2k} = \frac{a}{1-r^2}

and for odd powers only,

\sum_{k=0}^\infty ar^{2k+1} = \frac{ar}{1-r^2}

In cases where the sum does not start at k = 0,

\sum_{k=m}^\infty ar^k=\frac{ar^m}{1-r}

The formulae given above are valid only for |&nbsp;r&nbsp;|&nbsp;<&nbsp;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 p-adic numbers if |&nbsp;r&nbsp;|p&nbsp;<&nbsp;1. As in the case for a finite sum, we can differentiate to calculate

Arithmetic progression

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, &hellip; 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.


The sum of the members of a finite arithmetic progression is called an arithmetic series.

Expressing the arithmetic series in two different ways:


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 re-inserting the substitution: a_n = a_1 + (n-1)d:

S_n=\frac{n}{2}[ 2a_1 + (n-1)d].

In 499 CE Aryabhata, a prominent mathematician-astronomer 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 an = 3 + (n-1)(5) up to the 50th term is

S_{50} = \frac{50}{2}[2(3) + (49)(5)] = 6,275.


The product of the members of a finite arithmetic progression with an initial element a1, 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 (n-2) \times (n-1) \times n \,\!

for positive integers m and n is given by


Taking the example from above, the product of the terms of the arithmetic progression given by an = 3 + (n-1)(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


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^{n-1} d^{n-2} + (n-1)a^{n-2}d^{n-1}

so the product of n terms of an AP is:

a^{n} + na^{n-1} d^{n-2} + (n-1)a^{n-2}d^{n-1} no solutions

From Yahoo Answers

Question:I have no idea how to solve this. I dont want an answer to the problem, just instuctions or a formula for how to solve it myself. I have to show my work, so I dont want an answer. Find the sum of the first four terms of the geometric sequence with a = -3 and r = 3.

Answers:Sn = a1(1-r^n)/(1-r) 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*(1-3^4)/(1-3) S4 = -3*-80/-2 S4 = -120

Question:What is the equation you use to find the sum of a geometric series?

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 -----

Question:what is the least possible integer that can be the sum of an infinite geometric progression whose first term is 10? please explain the answer too, i do not really get the whole concept of geometric progression, i already tried wiki put i couldn't comprehend it. also this was an extra credit question I'm just curious and am only in geometry honors right now as a freshman, so please don't call me stupid... also i know the question has incorrect grammar, i couldn't fit all if i used correct spelling, etc.

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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