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a2=2 a3=10 in a geometric series. Find the common ratio & a1

a2=2 a3=10 in a geometric series. Find the common ratio & a1

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If a5/a3=4/9 and a2=4/9 find the nth geometric sequence.?


If a5/a3=4/9 and a2=4/9 find the nth geometric sequence.? Delete. ... \\~\\ a=\dfrac{2}{3}\\~\\~\\ \text{nth term}\\ =ar ... Find more explanations on ...
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A geometric sequence is defined by the general term tn = - page 5


A geometric sequence is defined by ... first four terms of a geometric sequence if the common ratio is 10 ... find a general term for the sequence a1,a2,a3,a4 ...
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11.3 – Geometric Sequences


... term r is the common ratio Geometric Mean Find the product ... a1 * r n-1 Now find the first five terms: a1 = 7(1/3) (1-1) = 7 a2 = 7(1/3) (2-1) = 7/3 a3 = 7(1 ...
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Biology : Determine the first four terms of the geometric ...


Biology : Determine the first four terms of the geometric sequence. a1= -7 r = -4, , ... For geometric ... (a1) is -7 and the common ratio (r) ... a1=2, a2=5, a3= 8, ...
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Arithmetic and Geometric Sequences - ppt download


... arithmetic and geometric ... common ratio is r, and the first term a1 is often referred to simply as "a". Since you get the next term by multiplying by the common ...
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A geometric sequence is defined by the general term tn = - page 4


Find the common ratio of the geometric sequences ... b) Using the formula for the sum of the first n terms of a geometric series, ... 3)/(2n − 1) a1 = ? a2 = ? ...
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Geometric Progression Essay - 378 Words


Geometric Progression and Double Bedrooms Essay ...… is a quadratic sequence. 2.1 Write down the next term.
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Assessment 0908 Geometric Series Essay - 1048 Words


Assessment 0908 Geometric Series . ... Series & Sums Introduction ... , where | r | common ratio | | a1 | first term | | a2 | second term ...
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Produced with a Trial Version of PDF Annotator - www ...


I= ∞ exp(−x2 )dx 2 3 Example 4.8 & ... a1 , a2 , a3 , . . .}, ... ∑ xn n=0 This is a geometric series with common ratio x.
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(a) an ordered list of objects. - ppt download - SlidePlayer


... an ordered list of objects. ... 1, 2, 3, 4, …,n… Range a1, a2, a3, ... 14 Find r (the common ratio) Geometric Series Each term is obtained from the preceding ...
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a2=2 a3=10 in a geometric series. Find the common ratio & a1

10/2=5 term 1=2/5=0.4 ..................
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circumference of an oval that is 100 inches by 72 inches

The oval described seems to be an ellipse, the equation of which is x^2/a^2+y^2/b^2=1 where a and b are the "radii", the semi-major and semi-minor axes. a=100/2=50 and b=72/2=36 so the equation is x^2/2500+y^2/1296=1. An approximate answer for the circumference is given by 2π√((a^2+b^2)/2)=273.3 in. METHOD USING CALCULUS To find the circumference we need calculus, or something that gives us a good approximation. First, find the eccentricity, e: b^2=a^2(1-e^2) so 1296=2500(1-e^2), 1-e^2=0.5184, e=√0.4816=0.6940 approx. An ellipse can be cut into 4 pieces of equal size so we only need the length of the arc of one to find the whole circumference. Consider a point P(x,y) on the ellipse. A small arc (segment of the circumference) is created as x and y change. If x changes by an amount dx and y by dy, then the arc ds=√(dx^2+dy^2) by Pythagoras. The circumference is the sum of all such ds or ∫ds. ds/dx=√(1+(dy/dx)^2), so ∫ds=∫(√(1+(dy/dx)^2)dx). x^2/a^2+y^2/b^2=1, x^2/a^2+y^2/(a^2(1-e^2))=1 (1) y^2/(a^2(1-e^2))=1-x^2/a^2=(a^2-x^2)/a^2 for all ellipses with centre at the origin (0,0). y^2=a^2(1-e^2)(a^2-x^2)/a^2=(1-e^2)(a^2-x^2), y=√((1-e^2)(a^2-x^2)). Differentiate (1): 2x/a^2+(2y/(a^2(1-e^2)))dy/dx=0; dy/dx=-x/a^2 * (a^2(1-e^2))/√((1-e^2)(a^2-x^2)); dy/dx=-x√((1-e^2)/(a^2-x^2)) and (dy/dx)^2=x^2(1-e^2)/(a^2-x^2). 1+(dy/dx)^2=1+x^2(1-e^2)/(a^2-x^2)=(a^2-x^2+x^2-e^2x^2)/(a^2-x^2)=(a^2-e^2x^2)/(a^2-x^2). s=∫ds=∫(√((a^2-e^2x^2)/(a^2-x^2))dx)=∫(√((1-e^2x^2/a^2)/(1-x^2/a^2))dx). [Let x=asin(t) then dx/dt=acos(t). s=∫((√(1-e^2sin^2(t))/cos(t))dx)=a∫(√(1-e^2sin^2(t))dt. The whole circumference is 4s=4a∫(√(1-e^2sin^2(t))dt. The lower limit for t is x/a when x=0, so t=0; the upper limit is x/a when x=a so t=90° or π/2 radians.] [a^2=2500, e^2=0.4816, s=∫ds=∫(√((1-0.4816x^2/2500)/(1-x^2/2500))dx)=∫√((1-0.00019264x^2)/(1-0.0004x^2))dx).] The limits of this integral for a quadrant of the ellipse are x=0 to x=a=50. Rather than attempting to integrate this directly, we may be able to reduce the expression under the square root to a series: By algebraic division, (1-e^2(x/a)^2)/(1-(x/a)^2)=1+(x/a)^2(1-e^2)+(x/a)^4(1-e^2)+(x/a)^6(1-e^2)+... Let z=(x/a)^2(1-e^2)+...+(x/a)^2n(1-e^2)+... =(1-e^2)((x/a)^2+(x/a)^4+(x/a)^6+...). This is a geometric progression with common ratio (x/a)^2.  So z/(1-e^2)=∑(x/a)^2i for 1≤i≤n. The sum, Sn, to n terms of A+Ar+Ar^2+...+Ar^(n-1), where A is the first term and r is the common ratio is found as follows: rSn=Ar+Ar^2+Ar^3+...+Ar^n, so rSn-Sn=Ar^n-A and Sn=A(r^n-1)/(r-1). Put A=R=(x/a)^2: z/(1-e^2)=(x/a)^2((x/a)^2n-1)/((x/a)^2-1). Unfortunately this series does not converge quickly because x/a starts off at 0 but finishes up as 1. Nevertheless we can attempt to use it as follows: s=∫ds=∫((1+z)^½dx)=∫((1+z/2-z^2/8+z^3/16-5z^4/128+...)dx)= ∫((1+(x/a)^2(1-e^2)/2+(x/a)^4(1-e^2)/2+...)-((x/a)^2(1-e^2)+(x/a)^4(1-e^2)+...)^2/8+...)dx). If we ignore all terms beyond (x/a)^4 we end up with ∫((1+(x/a)^2(1-e^2)/2+(x/a)^4(1-e^2)/2-(x/a)^4(1-e^2)^2/8)dx)= (x+x^3(1-e^2)/6a^2+x^5(1-e^2)/10a^4-x^5(1-e^2)^2/40a^4) for 0≤x≤50. When x=0, the lower limit, this expression is zero. This simplifies to a(1+(1-e^2)/6+(1-e^2)/10-(1-e^2)^2/40) when we put x=a, the upper limit. So putting a=50 and 1-e^2=0.5184 we have s=50(1+0.5184/6+0.5184/10-0.5184^2/40)= 50(1+0.0864+0.05184-0.006718)=56.576 inches. This is the arc of a quadrant so the circumference of the ellipse is approximately 4*56.576=226.3 inches. This is shorter than the approximate circumference (273.3") given at the beginning, so there appears to be an error. The error is that we haven't taken sufficient terms, so we need to look for a better approximation method.
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Find the 8th and 13th term of the finite geometric series: -1+2+(-4)+8+(-16)+...a20

Question: Find the 8th and 13th term of the finite geometric series: -1+2+(-4)+8+(-16)+...a20 A geometric sequence is given by an = a1*r^(n-1) where an is the nth term, a1 is the 1st term, r is the common ratio. Looking at your series: -1+2+(-4)+8+(-16)+...a20 a1 = -1 and r = -2   8th term a8 = a1*r^(8-1) = (-1)*(-2)^7 = (-1)*(-128) a8 = 128 13th term a13 = a1*r^(13-1) = (-1)*(-2)^(12) = (-1)*4096 a13 = -4096 p.s. if you wish to clarify that you are using a subscript form, then use an underscore. e.g. write a20 as a_20 If the context of the text you are writing about is sufficiently clear about the use of subscripts then a20 is fine.
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Find the explicit formula of the given sequence.

Arithmetic series Between a2 and a5 we have a3 and a4. If the common difference is d, then a3=a2+d, a4=a3+d, a5=a4+d, so a5-a3=3d, because a4=a3+d=a2+d+d=a2+2d, and a5=a4+d=a2+2d+d=a2+3d. We are given a2=5 and a5=-4. a5-a2=-9=3d, so d=-3. a1=a2+3=8, a0=a1+3=11. Now we can write a formula for a(n)=11-3n, where n starts from 0. Check: a2=11-6=5 and a5=-4. Geometric series a5=a2r^3, where r is the common ratio. So 600=75r^3 and r^3=600/75=8, so r=2. a1=a2/2=75/2=37.5 and a0=a1/2=18.75. The formula for a(n)=18.75*2^n. Check: a2=18.75*4=75, a5=18.75*32=600.
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Determine wheater the infinite geometric series has a finite sum.... -49+(-7)+(-1/7)+...

1. The first term, a, is -49 and the common ratio, r, is 1/7. The sum, S, to n terms is S=a(1-r^n)/(1-r). When n is very large  r^n gets very close to zero, so S=a/(1-r)=-49/(6/7)=-49*7/6=-343/6 or -57.1666...7. This is the finite value  which the infinite series approaches. 2. You seem to be describing a summation symbol (capital Greek letter sigma, resembling E) between limits for i between 1 and infinity (symbol is like an 8 on its side). This is a shorthand way of writing an infinite geometric series where the general term is 5^(i/2). The first term is 5^(1/2)=sqrt(5) and the common ratio is the same as the first term. The common ratio is bigger than 1 because sqrt(5) is bigger than 2, therefore the series doesn't converge to a finite value, and its sum would be infinite.
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an geometric sequences is given as (x+40),(x+4),(x-20),...find the common ratio,r

3 points...x+40, x+4, x-20 distans tween points: 36, 24 24/36=2/3 maebee distans tween terms=(2/3)*prior term
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arithmetic sequence 1;x;y;... results in geometric sequence if each term is decreased by 1

I assume each term does not include the first term 1, otherwise there could be no GP since all terms would be zero. AP: terms are 1, 1+p, 1+2p, 1+3p, etc. GP: terms are 1, r, r^2, r^3, etc. p is the common difference in the AP and r the common ratio in the GP. x=1+p and y=1+2p, so p=y-x. x-1=r and y-1=r^2. Since p=y-x and p=x-1, y-x=x-1 and y=2x-1. y-1=(x-1)^2 because r=x-1, so y=x^2-2x+2=2x-1. Therefore x^2-4x+3=0 and (x-3)(x-1)=0, making x=1 or 3, and making y=1 and r=p=0; or y=5 and r=p=2. The AP cannot have a zero common difference, and the GP cannot have a zero common ratio, so we accept the set of values x=3, y=5 and r=p=2. The AP becomes 1, 3, 5, 7, etc., and the GP becomes 1, 2, 4, 8, etc. The nth term of the AP is 2n-1 and of the GP 2^(n-1).  
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Please see full question below, what is the first term of the series? See question below

If a is the first term then the series runs a+a/4+a/16+a/64+.. until the fraction becomes infinitesimally small. But let's not take it that far. We'll go as far as n. Let S=sum of the series up to n, so the last term in the series is a/4^n. Now consider multiplying the whole series by 1/4, so we get S/4=a/4+a/16+a/64+a/256+...+a/4^(n+1). Now subtract this new series from S and we get 3S/4=a-a/4^(n+1). As n approaches infinity a/4^(n+1) approaches zero, so we can drop the term and 3S/4=a. We know the converging value of the series S=20/3, therefore a=(3/4)*(20/3)=5. The first term is 5, the next term is 5/4, and so on.
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Geometrical series

The constant 12 can be taken outside the summation because it applies to every term. The series becomes: 12(-1/5+1/25-1/125+1/625-...). The series is a GP with common factor r=-1/5. the series can be written -(12/5)(1-1/5+1/25-...). Call the series in brackets S. So S=1-1/5+1/25-...+5^-n. Therefore S/5=1/5-1/25+...+5^-(n+1) and S+S/5=1+5^-(n+1). When n is very large 5^-(n+1) becomes very small, and it doesn't matter whether it's positive or negative, so we can say that as n approaches infinity the term becomes 0. Therefore S+S/5=6S/5=1 and S=5/6. Returning to the original GP, we have -(12/5)S=-12/5*5/6=-2. So the series converges to -2.  
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Find The sum of the infinite series in which the first term is 3/2 and the common ratio is 1/4.

??? yu want ??? 1.5+0.375+0.09375+-.0234375 +0.005859+0.0014658 sum=2 =1.5+0.5
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