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3 years ago

Find all geometric sequences such that the sum of the first two terms is 24 and the sum of the first three terms is 26.

Mathematics

1 answer:

Serga [27]3 years ago

6 0

Answer:

<h3>The nth term Tn = -8(-1/4)^(n-1) or Tn = 6(1/3)^(n-1) can be used to find all geometric sequences</h3>

Step-by-step explanation:

Let the first three terms be a/r, a, ar... where a is the first term and r is the common ratio of the geometric sequence.

If the sum of the first two term is 24, then a/r + a = 24...(1)

and the sum of the first three terms is 26.. then a/r+a+ar = 26...(2)

Substtituting equation 1 into 2 we have;

24+ar = 26

ar = 2

a = 2/r ...(3)

Substituting a = 2/r into equation 1 will give;

(2/r))/r+2/r = 24

2/r²+2/r = 24

(2+2r)/r² = 24

2+2r = 24r²

1+r = 12r²

12r²-r-1 = 0

12r²-4r+3r -1 = 0

4r(3r-1)+1(3r-1) = 0

(4r+1)(3r-1) = 0

r = -1/4 0r 1/3

Since a= 2/r then a = 2/(-1/4)or a = 2/(1/3)

a = -8 or 6

All the geometric sequence can be found by simply knowing the formula for heir nth term. nth term of a geometric sequence is expressed as

if r = -1/4 and a = -8

Tn = -8(-1/4)^(n-1)

if r = 1/3 and a = 6

Tn = 6(1/3)^(n-1)

The nth term of the sequence above can be used to find all the geometric sequence where n is the number of terms

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Given:

Height of man = 6 ft

Height of man's shadow = 11 feet

Height of building's shadow = 139 feet

To find:

The height of the building.

Solution:

We know that the heights of the objects and there shadows are always proportional.

$\dfrac{\text{Height of man}}{\text{Height of man's shadow}}=\dfrac{\text{Height of the building}}{\text{Height of building's shadow}}$

Let x be the height of the building.

$\dfrac{6}{11}=\dfrac{x}{139}$

Multiply both sides by 139.

$\dfrac{6}{11}\times 139=x$

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2 years ago

You must design a closed rectangular box of width w, length l and height h, whose volume is 504 cm3. The sides of the box cost 3

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Answer:

Dimensions will be

Length = 7.23 cm

Width = 7.23 cm

Height = 9.64 cm

Step-by-step explanation:

A closed box has length = l cm

width of the box = w cm

height of the box = h cm

Volume of the rectangular box = lwh

504 = lwh

$h=\frac{504}{lw}$

Sides which involve length and width and height, cost = 3 cents per cm²

Top and bottom of the box costs = 4 cents per cm²

Cost of the sides $C_{s}$ = 3[2(l + w)h] = 6(l + w)h

$C_{s}$ = 3[2(l + w)h]

$C_{s}=6(l+w)(\frac{504}{lw} )$

Cost of the top and the bottom $C_{(t,p)}$ = 4(2lw) = 8lw

Total cost of the box C = $3024\frac{(l+w)}{lw}$ + 8lw

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To minimize the cost of the sides

$\frac{dC}{dl}=3024(-l^{-2}+0)+8w=0$

$\frac{3024}{l^{2}}=8w$

$\frac{378}{l^{2}}=w$ ---------(1)

$\frac{dC}{dw}=3024(-w^{-2})+8l=0$

$\frac{3024}{w^{2}}=8l$

$\frac{378}{w^{2}}=l$

$w^{2}=\frac{378}{l}$ -------(2)

Now place the value of w from equation (1) to equation (2)

$(\frac{378}{l^{2}})^{2}=\frac{378}{l}$

$\frac{(378)^{2} }{l^{4}}=\frac{378}{l}$

l³ = 378

l = ∛378 = 7.23 cm

From equation (2)

$w^{2}=\frac{378}{7.23}$

$w^{2}=52.28$

w = 7.23 cm

As lwh = 504 cm³

(7.23)²h = 504

$h=\frac{504}{(7.23)^{2}}$

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