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3 years ago

The common ratio of a geometric series is 1/4 and the sum of the first 4 terms is 170.

Mathematics

1 answer:

Illusion [34]3 years ago

5 0

Answer: the first term of the series is 128

Step-by-step explanation:

In a geometric sequence, the consecutive terms differ by a common ratio. The formula for determining the sum of n terms, Sn of a geometric sequence is expressed as

Sn = a(1 - r^n)/(1 - r)

Where

n represents the number of term in the sequence.

a represents the first term in the sequence.

r represents the common ratio.

From the information given,

r = 1/4 = 0.25

n = 4

S4 = 170

Therefore, the expression for the sum of the 4 terms, S4 is

170 = a(1 - 0.25^4)/(1 - 0.25)

170 = a(1 - 0.00390625)/(1 - 0.25)

170 = a(0.99609375)/(0.75)

170 = 1.328125a

a = 170/1.328125

a = 128

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Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the le

Svet_ta [14]

Answer:

The maximum possible length, in centimeters, of the longest piece of rope is 134 cm

Step-by-step explanation:

Let the seven pieces be a,b,c,d,e,f,g

Average of 7 pieces of ropes =68

$Average = \frac{\text{Sum of all lengths}}{\text{No. of pieces}}$

$68= \frac{\text{Sum of all lengths}}{7}$

$68 \times 7 =\text{a+b+c+d+e+f+g}$

$476=\text{a+b+c+d+e+f+g}$ --A

We are given that the median length of a piece of rope is 84 centimeters.

We arrange the pieces in the ascending order

So, median will be the length of 4th piece

So, d = 84 cm

the length of piece a,b,c will be less than 84 and the value of e,f,g must be greater than or equal to 84

Let us suppose the length of a,b,c be x

Let us suppose the length of e,f be 84

The length of the longest piece of rope is 14 centimeters more than 4 times the length of the shortest piece of rope

So, g = 4x+14

Substitute the value in A

$476=x+x+x+84+84+84+4x+14$

$476=7x+266$

$\frac{476-266}{7}=x$

$30=x$

g = 4x+14=4(30)+14=134

Hence the maximum possible length, in centimeters, of the longest piece of rope is 134 cm

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and the plane

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To see clearly that this is an ellipse, le us divide through by 16, to get

$\frac{x^2}{1}+ \frac{y^2}{16}=1$

$\frac{x^2}{1^2}+ \frac{y^2}{4^2}=1$ ,

We can write the following parametric equations,

$x=cos(t), y=4sin(t)$

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Since C lies on the plane,

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Let us make z the subject first,

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