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4.8: Planes in Rⁿ - Mathematics


Planes in (mathbb{R}^n)

Learning Objectives

  • Find the vector and scalar equations of a plane.

Much like the above discussion with lines, vectors can be used to determine planes in (mathbb{R}^n). Given a vector (vec{n}) in (mathbb{R}^n) and a point (P_0), it is possible to find a unique plane which contains (P_0) and is perpendicular to the given vector.

Definition (PageIndex{1}): Normal Vector

Let (vec{n}) be a nonzero vector in (mathbb{R}^n). Then (vec{n}) is called a normal vector to a plane if and only if [vec{n} ullet vec{v} = 0] for every vector (vec{v}) in the plane.

In other words, we say that (vec{n}) is orthogonal (perpendicular) to every vector in the plane.

Consider now a plane with normal vector given by (vec{n}), and containing a point (P_0). Notice that this plane is unique. If (P) is an arbitrary point on this plane, then by definition the normal vector is orthogonal to the vector between (P_0) and (P). Letting (overrightarrow{0P}) and (overrightarrow{0P_0}) be the position vectors of points (P) and (P_0) respectively, it follows that [vec{n} ullet (overrightarrow{0P} - overrightarrow{0P_0}) = 0] or [vec{n} ullet overrightarrow{P_0P} = 0]

The first of these equations gives the vector equation of the plane.

Definition (PageIndex{2}): Vector Equation of a Plane

Let (vec{n}) be the normal vector for a plane which contains a point (P_0). If (P) is an arbitrary point on this plane, then the vector equation of the plane is given by [vec{n} ullet (overrightarrow{0P} - overrightarrow{0P_0}) = 0]

Notice that this equation can be used to determine if a point (P) is contained in a certain plane.

Example (PageIndex{1}): A Point in a Plane

Let (vec{n} = left[ egin{array}{r} 1 2 3 end{array} ight]) be the normal vector for a plane which contains the point (P_0 = left( 2, 1, 4 ight)). Determine if the point (P = left( 5, 4, 1 ight)) is contained in this plane.

Solution

By Definition [def:vecteqnplane], (P) is a point in the plane if it satisfies the equation [vec{n} ullet (overrightarrow{0P} - overrightarrow{0P_0}) = 0]

Given the above (vec{n}), (P_0), and (P), this equation becomes [egin{aligned} left[ egin{array}{r} 1 2 3 end{array} ight] ullet left( left[ egin{array}{r} 5 4 1 end{array} ight] - left[ egin{array}{r} 2 1 4 end{array} ight] ight) &=& left[ egin{array}{r} 1 2 3 end{array} ight] ullet left( left[ egin{array}{r} 3 3 -3 end{array} ight] ight) &=& 3 + 6 - 9 = 0end{aligned}]

Therefore (P = ( 5, 4, 1)) is contained in the plane.

Suppose (vec{n} = left[ egin{array}{c} a b c end{array} ight]), (P = left( x,y,z ight)) and (P_0 = (x_0, y_0, z_0 )).

Then [egin{aligned} vec{n} ullet (overrightarrow{0P} - overrightarrow{0P_0}) &=& 0 left[ egin{array}{c} a b c end{array} ight] ullet left( left[ egin{array}{c} x y z end{array} ight] - left[ egin{array}{c} x_0 y_0 z_0 end{array} ight] ight) &=& 0 left[ egin{array}{c} a b c end{array} ight] ullet left[ egin{array}{c} x - x_0 y - y_0 z - z_0 end{array} ight] &=& 0 a(x - x_0) + b (y - y_0) + c (z-z_0) &=& 0 end{aligned}]

We can also write this equation as [ax + by + cz = ax_0 + by_0 + cz_0]

Notice that since (P_0) is given, (ax_0+by_0+cz_0) is a known scalar, which we can call (d). This equation becomes [ax + by + cz = d]

Definition (PageIndex{3}): Scalar Equation of a Plane

Let (vec{n} = left[ egin{array}{c} a b c end{array} ight]) be the normal vector for a plane which contains the point (P_0 = (x_0, y_0, z_0)).Then if (P=(x,y,z)) is an arbitrary point on the plane, the scalar equation of the plane is given by [ax + by + cz = d] where (a,b,c,d in mathbb{R}) and (d = ax_0 + by_0 + cz_0).

Consider the following equation.

Example (PageIndex{1}): Finding the Equation of a Plane

Find an equation of the plane containing (P_0 = (3, -2, 5)) and orthogonal to (vec{n} = left[ egin{array}{r} -2 4 1 end{array} ight]).

Solution

The above vector (vec{n}) is the normal vector for this plane. Using Definition [def:vecteqnplane], we can determine the vector equation for this plane. [egin{aligned} vec{n} ullet (overrightarrow{0P} - overrightarrow{0P_0}) &=& 0 left[ egin{array}{r} -2 4 1 end{array} ight] ullet left(left[ egin{array}{c} x y z end{array} ight] - left[ egin{array}{r} 3 -2 5 end{array} ight] ight) &=& 0 left[ egin{array}{r} -2 4 1 end{array} ight] ullet left[ egin{array}{c} x - 3 y + 2 z - 5 end{array} ight] &=& 0 end{aligned}]

Using Definition [def:scalareqnplane], we can determine the scalar equation of the plane. [-2x + 4y + 1z = -2(3) + 4(-2) + 1(5) = -9]

Hence, the vector equation of the plane is [left[ egin{array}{r} -2 4 1 end{array} ight] ullet left[ egin{array}{c} x - 3 y + 2 z - 5 end{array} ight] = 0] and the scalar equation is [-2x + 4y + 1z = -9]

Suppose a point (P) is not contained in a given plane. We are then interested in the shortest distance from that point (P) to the given plane. Consider the following example.

Example (PageIndex{1}): Shortest Distance From a Point to a Plane

Find the shortest distance from the point (P = (3,2,3)) to the plane given by
(2x + y + 2z = 2), and find the point (Q) on the plane that is closest to (P).

Solution

Pick an arbitrary point (P_0) on the plane. Then, it follows that [overrightarrow{QP} = func{proj}_{vec{n}}overrightarrow{P_0P}] and (| overrightarrow{QP} |) is the shortest distance from (P) to the plane. Further, the vector (overrightarrow{0Q} = overrightarrow{0P} - overrightarrow{QP}) gives the necessary point (Q).

From the above scalar equation, we have that (vec{n} = left[ egin{array}{c} 2 1 2 end{array} ight]). Now, choose (P_0 = (1, 0, 0)) so that (vec{n} ullet overrightarrow{0P} = 2 = d). Then, (overrightarrow{P_0P} = left[ egin{array}{c} 3 2 3 end{array} ight] - left[ egin{array}{c} 1 0 0 end{array} ight] = left[ egin{array}{c} 2 2 3 end{array} ight]).

Next, compute (overrightarrow{QP} = func{proj}_{vec{n}}overrightarrow{P_0P}). [egin{aligned} overrightarrow{QP} &=& func{proj}_{vec{n}}overrightarrow{P_0P} &=& left( frac{ overrightarrow{P_0P} ullet vec{n}}{| vec{n} | ^2} ight)vec{n} &=& frac{12}{9} left[ egin{array}{r} 2 1 2 end{array} ight] &=& frac{4}{3} left[ egin{array}{r} 2 1 2 end{array} ight] end{aligned}]

Then, (| overrightarrow{QP} | = 4) so the shortest distance from (P) to the plane is (4).

Next, to find the point (Q) on the plane which is closest to (P) we have [egin{aligned} overrightarrow{0Q} &=& overrightarrow{0P} - overrightarrow{QP} &=& left[ egin{array}{r} 3 2 3 end{array} ight] - frac{4}{3} left[ egin{array}{r} 2 1 2 end{array} ight] &=& frac{1}{3} left[ egin{array}{r} 1 2 1 end{array} ight]end{aligned}]

Therefore, (Q = (frac{1}{3}, frac{2}{3}, frac{1}{3} )).


AffineSpaces.jl

This library provides a simple, general way of dealing with affine subspaces, and it's implemented entirely in Julia. Affine subspaces are familiar objects like points, lines, planes, etc., as well as more exotic objects like hyperplanes and so on. Using this library, you can represent and manipulate objects like this easily.

What is it good for?

Often we want to do things like calculate plane-line intersections, or plane-plane intersections, or find the plane that passes through a line and a point, or find the distance from a point to a line in 3d, and lots of other geometric operations. There are countless Q&As on stackoverflow dealing with each of these special cases. But unforunately the code is usually very specific and not generalizable. It's also not guaranteed to work for all special cases (if planes are parallel, for example) and might be numerically unstable. For example, the intersection of three planes (in 3d) might return a point, a line, a plane, or nothing at all, and our code needs to be able to reflect this.

This kind of complexity and confusion is where this library comes in: You can do all these computational geometry operations and more in a way that's guaranteed to be mathematically correct. And it's all done through a unified interface - no special cases.

So how do I use it?

First, let's go through what an affine subspace is. An affine subspace is built upon the notion of a vector space. In mathematics, vector spaces can be defined in a lot of different ways, but here we're going to stick to a very simple way of defining them. We're simply going to define vector spaces as linear combinations of basis vectors. So for example, if you have a basis vector [1,0,0] , the vector space generated by this basis vector includes all vectors a*[1,0,0] where a is a real number, so it would be all vectors of the form [a,0,0] . If you have two basis vectors, say [1,0,0] (as before) and [1/2,1,0] , now your vector space consists of all vectors a*[1,0,0] + b*[1/2,1,0] , where both a and b are real numbers. So they would look like: [a+b/2,b,0] . In AffineSpaces.jl, we would define this vector space as follows:

That is, by giving the basis matrix as the argument to the constructor.

You can see that by manipulating basis vectors, you can generate lines (one basis vector), planes (two basis vectors), and so on.

Affine Subspaces

Affine subspaces are simply vector spaces plus some offset. The vector spaces we defined had to pass through the origin - they all had to have [0,0,0] in the space. This puts a limitation on the kinds of lines and planes we can represent. Affine subspaces don't need to pass through the origin. For example, here is the line y=1 in 2d:

You can show that with this simple formalism, any kind of point, line, plane, hyperplane, etc. can be represented.

Operations

Distance

In AffineSpaces.jl, all affine subspace operations can be done between any two affine spaces. The distance between two affine spaces, for example, can be calculated easily:

And this works no matter what the two affine subspaces are. You can use it to calculate, for example, the distance between two points, a plane and a point, or two parallel planes. The only restriction is that the two subspaces inhabit the same space i.e. they both inhabit 2d or 3d space. But that's it. If the two affine subspaces intersect, the distance returned will simply be zero.

The fact that this function is so general is even more interesting when you look at the implementation of the function:

This is the power of working with a general affine subspace structure - very general calculations can be performed in a simple way.

Generated subspaces

generated_space is the function that takes two affine subspaces and produces the smallest affine subspace that includes both. So for instance, we can calculate the line that passes through two points:

If the two points are coincident, it won't throw an error - it will just return the point, as it should.

Another example: the plane that passes through a point and a line:

Again, if the point lies on the line, it will just return the line. If line isn't a line, as we think, but is instead a point (for example, if it was returned from a previous operation), then it will return the line that goes through both points. In this way, in AffineSpaces.jl you don't have to worry about what your geometric objects are and you don't need to worry about writing special case code for them. Everything 'just works'.

Intersections

The following function can be used to calculate affine subspace intersections. The intersection is the largest affine subspace that is included in both.

Solid Geometry

AffineSpaces.jl contains a set of functions for performing solid geometry - the geometry of volumes including n-dimensional polyhedra. These are described below.

Half-spaces

The simplest volume is the entire R n space. We can represent this space using basis vectors, as mentioned. However, R n by itself is not that interesting. It becomes more interesting when we introduce the notion of half-spaces. Half-spaces are produced when we take an n-dimensional space and divide it into parts using a hyperplane (an affine subspace of dimension n-1). For example, we can divide the 2D plane into two parts with a line, or a 3D space into two parts with a plane. Mathematically, we can represent a half-space of dimension n as follows. Let a be a normal vector of dimension n, and b be a real number. Then all points x in the space that satisfy:

a T x > b

Or, if it's a closed half-space:

a T x ≥ b

Together form a half-space. To create a half-space in AffineSpaces.jl, we simply input a and b, for example here's a 3D half-space:

The first parameter is the normal vector, the second parameter is the offset, and the final parameter specifies if the half-space is closed or not.

Convex polyhedra

Now that we have a notion of half-spaces, we can combine them together via intersections, using the function inter() . Half-spaces are convex spaces. The intersection of any two convex spaces is also a convex space, so the intersection of any number of half-spaces is also a convex space. Indeed, you can show that any convex space/polyhedron can be formed via an intersection of half-spaces. This idea forms the basis of Nef polyhedron theory, and gives us a simple mathematical formalism for dealing with convex polyhedra.


M4TH 22I

∴ can get all soln to Ax=b from any one sol to Ax=b (ex. p) and adding it to q, soln set of Ax=0.

easy case - triangular matrix
(a₁₁ -------
0 a₂₂ ______
0 0 a₃₃ ___)

expansion in first row:
det(A) = a₁₁det(A₁₁) - 0det(A₁₂) +0det(A₁₃)
a₁₁det(n-1) = a₁₁a₂₂det(n-2) =a₁₁a₂₂a₃₃.. =product of pivots

B=basis of Rⁿ =
r= (c₁. cn col) wrt B by (1)
Ar = (λ₁c₁. λncn col) wrt B by (2)

pick b₀ in col(A) closest to b and solve Ax=b₀ instead.

or w₂ is best approx to x in W⊥ and w₁ is error

given any A there is one and only one X which will satisfy the equation.

Check the true statements below:

A. If det(A) is zero, then two rows or two columns are the same, or a row or a column is zero.

B. If two row interchanges are made in sucession, then the determinant of the new matrix is equal to the determinant of the original matrix.


A Mathematician's View of Evolution

When Dr. Behe was at the University of Texas El Paso in May of 1997 to give an invited talk, I told him that I thought he would find more support for his ideas in mathematics, physics and computer science departments than in his own field. I know a good many mathematicians, physicists and computer scientists who, like me, are appalled that Darwin's explanation for the development of life is so widely accepted in the life sciences. Few of them ever speak out or write on this issue, however--perhaps because they feel the question is simply out of their domain. However, I believe there are two central arguments against Darwinism, and both seem to be most readily appreciated by those in the more mathematical sciences.

    The cornerstone of Darwinism is the idea that major (complex) improvements can be built up through many minor improvements that the new organs and new systems of organs which gave rise to new orders, classes and phyla developed gradually, through many very minor improvements. We should first note that the fossil record does not support this idea, for example, Harvard paleontologist George Gaylord Simpson ["The History of Life," in Volume I of "Evolution after Darwin," University of Chicago Press, 1960] writes:

If a billion engineers were to type at the rate of one random character per second, there is virtually no chance that any one of them would, given the 4.5 billion year age of the Earth to work on it, accidentally duplicate a given 20-character improvement. Thus our engineer cannot count on making any major improvements through chance alone. But could he not perhaps make progress through the accumulation of very small improvements? The Darwinist would presumably say, yes, but to anyone who has had minimal programming experience this idea is equally implausible. Major improvements to a computer program often require the addition or modification of hundreds of interdependent lines, no one of which makes any sense, or results in any improvement, when added by itself. Even the smallest improvements usually require adding several new lines. It is conceivable that a programmer unable to look ahead more than 5 or 6 characters at a time might be able to make some very slight improvements to a computer program, but it is inconceivable that he could design anything sophisticated without the ability to plan far ahead and to guide his changes toward that plan.

If archeologists of some future society were to unearth the many versions of my PDE solver, PDE2D , which I have produced over the last 20 years, they would certainly note a steady increase in complexity over time, and they would see many obvious similarities between each new version and the previous one. In the beginning it was only able to solve a single linear, steady-state, 2D equation in a polygonal region. Since then, PDE2D has developed many new abilities: it now solves nonlinear problems, time-dependent and eigenvalue problems, systems of simultaneous equations, and it now handles general curved 2D regions. Over the years, many new types of graphical output capabilities have evolved, and in 1991 it developed an interactive preprocessor, and more recently PDE2D has adapted to 3D and 1D problems. An archeologist attempting to explain the evolution of this computer program in terms of many tiny improvements might be puzzled to find that each of these major advances (new classes or phyla??) appeared suddenly in new versions for example, the ability to solve 3D problems first appeared in version 4.0. Less major improvements (new families or orders??) appeared suddenly in new subversions, for example, the ability to solve 3D problems with periodic boundary conditions first appeared in version 5.6. In fact, the record of PDE2D's development would be similar to the fossil record, with large gaps where major new features appeared, and smaller gaps where minor ones appeared. That is because the multitude of intermediate programs between versions or subversions which the archeologist might expect to find never existed, because-- for example--none of the changes I made for edition 4.0 made any sense, or provided PDE2D any advantage whatever in solving 3D problems (or anything else) until hundreds of lines had been added.

Whether at the microscopic or macroscopic level, major, complex, evolutionary advances, involving new features (as opposed to minor, quantitative changes such as an increase in the length of the giraffe's neck, or the darkening of the wings of a moth, which clearly could occur gradually) also involve the addition of many interrelated and interdependent pieces. These complex advances, like those made to computer programs, are not always "irreducibly complex"--sometimes there are intermediate useful stages. But just as major improvements to a computer program cannot be made 5 or 6 characters at a time, certainly no major evolutionary advance is reducible to a chain of tiny improvements, each small enough to be bridged by a single random mutation.

The biologist studies the details of natural history, and when he looks at the similarities between two species of butterflies, he is understandably reluctant to attribute the small differences to the supernatural. But the mathematician or physicist is likely to take the broader view. I imagine visiting the Earth when it was young and returning now to find highways with automobiles on them, airports with jet airplanes, and tall buildings full of complicated equipment, such as televisions, telephones and computers. Then I imagine the construction of a gigantic computer model which starts with the initial conditions on Earth 4 billion years ago and tries to simulate the effects that the four known forces of physics (the gravitational, electromagnetic and strong and weak nuclear forces) would have on every atom and every subatomic particle on our planet (perhaps using random number generators to model quantum uncertainties!). If we ran such a simulation out to the present day, would it predict that the basic forces of Nature would reorganize the basic particles of Nature into libraries full of encyclopedias, science texts and novels, nuclear power plants, aircraft carriers with supersonic jets parked on deck, and computers connected to laser printers, CRTs and keyboards? If we graphically displayed the positions of the atoms at the end of the simulation, would we find that cars and trucks had formed, or that supercomputers had arisen? Certainly we would not, and I do not believe that adding sunlight to the model would help much. Clearly something extremely improbable has happened here on our planet, with the origin and development of life, and especially with the development of human consciousness and creativity.

1 An unfortunate choice of words. I should have said, the underlying principle behind the second law is that natural forces do not do macroscopically describable things which are extremely improbable from the microscopic point of view. See my 2017 Physics Essays article, "On 'Compensating' Entropy Decreases," or the video above, for a more complete treatment of this point.


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**INRIA – Institut National de Recherche en Informatique et en Automatique (France)
***CMAME – Computer Methods in Applied Mechanics and Engineering


On a coordinate plane, a line goes through (negative 8, 4) and (8, 4). A point is at (negative 4, negative 6). What is the equation of the line that is parallel to the given line and passes through the point (−4,−6 )?

Since the y- coordinates are equal, this indicates the line is horizontal and parallel to the x- axis.

The equation of a horizontal line is

where c is the value of the y- coordinates the line passes through.

A parallel line will therefore be a horizontal line.

The line passes through (- 4, - 6 ) with y- coordinate - 6, thus

Equation of parallel line is y = - 6

a line named D goes through (-8, 4) and (8, 4) is parallel to the x-axis

then any other line parallel to line D must be parallel to the the x-axis too

then it has an equation of the form : y = a

and since it goes through (−4,−6 ) then its equation is : y= -6

The equation of parallel line passes through the point (−4,−6 ) is:

Please also check the attached graph.

The slope-intercept form of the line equation

m is the slopeb is the y-intercept

Given that on a coordinate plane passes through (-8, 4) and (8, 4).

As the value of y remains the same for any value of x.

i.e. y = 4 for any value of x.

Thus, the given line is a horizontal line.

We know that the slope of a horizontal line is zero.

Also it is clear that at x = 0, the value of y = 4. In other words, we can observe that the y-intercept b = 4.

so substitutig m = 0 and b = 4 in the slope-intercept form

Therefore, the equation of line passing through the point (-8, 4) and (8, 4) is:

Now, we also know that the slope of parallel lines is equal.

Thus, the slope of the line parallel to the line y = 4 is also 0.

As the parallel line passes through (−4,−6 ).

so substituting m = 0 and (-4, -6) in the slope-intercept form of line equation to determine the y-intercept b

Thus, the slope-intercept b = -6

now substituting m = 0 and b = -6 in the slope-intercept form of line equation

Therefore, the equation of parallel line passes through the point (−4,−6 ) is:


Unit 2 — Multivariable

1 4
2 3
. Show that
the image of the curve C under T is the circle x² + y² = 10.

The curve C is an ellipse in R2 (this follows from part (b), but you may assume this without proof). Find the area enclosed by C.

Let H : R2 → R2 be defined by H(x, y) = G(G(G(x, y))).

Use linear approximation to estimate H(0.95, 0.03).

At (0, 0), the direction of fastest increase of f + g is a positive scalar multiple of [1 -1]

Maybe! The direction of fastest increase is the sum of the gradients of f, g which is of the
form [ a −b] for a, b both positive. This could be
1
−1
but might not be.

Maybe. If the gradient of f + g is [ a −b], its directional derivative in [ 1 1] direction is a − b. Could be positive, could be negative

determinant of that matrix is equal to
0.1 * det of the lower bottom right square. Answer is 0.675.

A spider starts walking from the point (x, y) = (2, 3) with a velocity in the direction
1
−1

(centimeters per
second), so that its (x, y) position after t seconds is (2 + t, 3 − t). At what rate is its current temperature
changing as it leaves the starting point? (Related issue to be careful about here: when you are finding a
directional derivative, what difference does it make if you use a unit direction vector as opposed to one of
arbitrary length?)

, has a direct influence on its
heating/cooling rate that we do not want to ignore by dividing v by its magnitude.


Contents

Therefore, any totally regular summation method gives a sum of infinity, including the Cesàro sum and Abel sum. [1] On the other hand, there is at least one generally useful method that sums 1 + 2 + 4 + 8 + ⋯ to the finite value of −1. The associated power series

An almost identical approach (the one taken by Euler himself) is to consider the power series whose coefficients are all 1, that is,

The above manipulation might be called on to produce −1 outside the context of a sufficiently powerful summation procedure. For the most well-known and straightforward sum concepts, including the fundamental convergent one, it is absurd that a series of positive terms could have a negative value. A similar phenomenon occurs with the divergent geometric series 1 − 1 + 1 − 1 + ⋯ (Grandi's series), where a series of integers appears to have the non-integer sum 1 2 . <2>>.> These examples illustrate the potential danger in applying similar arguments to the series implied by such recurring decimals as 0.111 … and most notably 0.999 … . The arguments are ultimately justified for these convergent series, implying that 0.111 … = 1 9 <9>>> and 0.999 … = 1 , but the underlying proofs demand careful thinking about the interpretation of endless sums. [4]

It is also possible to view this series as convergent in a number system different from the real numbers, namely, the 2-adic numbers. As a series of 2-adic numbers this series converges to the same sum, −1, as was derived above by analytic continuation. [5]


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What is the sum of the geometric sequence 1, −6, 36, … if there are 6 terms?

By looking at the geometric sequence, we can note that each term is multiplied by -6.

So, the fourth term is going to be: -216

The fifth term is going to be: 1296

The sixth term is going to be: -7776

The sum of the geometric sequence is: 1 - 6 + 36 - 216 + 1296 - 7776 = -6665

We require the sum of 7 terms of the G.S. -1, 6, -36

The common ratio r = 6/-1 = -6 and the first term = -1.

Sum of n terms = a1 . (r^n - 1) / (r - 1)

Sum of 7 terms = -1 * (-6)^7 - 1) / (-6 - 1)

The required sum is 39991.

Step-by-step explanation: We are given to find the sum of the following 7-term geometric sequence:

Here, the first term , a = 1

common ratio, 'r' is given by

Thus, the required sum is 39991.

The correct answer option is A) -39,991.

We know that the sum of the geometric sequence is given by the formula:

where is the first term, is the common ratio and is the number of terms.

Substituting these values in the above formula to get:

The sum of a geometric series is given by:

Sn is the sum of the first n terms, r is the rate and n is the number of terms.

r the quotient between any two consecutive numbers.

Sum of the first six terms=6665

Sum of geometric sequence formula is


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