New questions (1 Viewer)

[currysauce]

If 2^x = 5^y = 10^z , using logs to base 10, or otherwise, show that 1/z = 1/x + 1/y

AND

In an arithmetic progression, whose first term and common difference are both non-zero, Un denotes the nth term and Sn denotes the sum of n terms. If U6, U4, U10 for a geometric progression:

i) show that S10=0
ii) show that S6 + S12 = 0
iii) deduce that U7 + U8 + U9 + U10 = U11 +U12

 

[Antwan23q]

are u sure that they are all equal to eachother?

 

[currysauce]

100% positive

 

[Antwan23q]

ok then

Log 2^x = Log 5^y = Log 10^z
xlog2 = ylog5 = z

so then 1) Log 5 = z/y
2) xlog10-xlog5 =z

x-xlog5= z
x-xz/y = z /z
x/z-x/y = 1 /x
1/z - 1/y = 1/x
.: 1/z = 1/x +1/y

 

[currysauce]

i can't follow ur working

 

[Antwan23q]

i can't follow ur working

lol, my bad mate, lemme rewrite it then.

Log 2^x = Log 5^y = Log 10^z
xlog2 = ylog5 = z

so then you get two equations,
1) ylog5 = z
2) xlog2 = z

but log 2 = log 10 - log 5
so then equation 2 becomes

xlog 10 - xlog 5 =z

x -xlog5 = z

but from equation one, we can say : log5 = z/y
so we sub that in,
x-xzy = z /then divide both sides by z
x/z - x/y = 1 /then divide both sides by x
1/z - 1/y = 1/x / then add 1/y to both sides
1/z =1/x +1/y

still workin on the second one

 

[currysauce]

but log 2 = log 10 - log 5
so then equation 2 becomes

xlog 10 - xlog 5 =z

x -xlog5 = z

ok i get ur working, except this, why do u shove x's infront of the log10 and log5, if they are absent from the equation above it

 

[香港!]

ok i get ur working, except this, why do u shove x's infront of the log10 and log5, if they are absent from the equation above it

xlog2=z
x(log10-log5)=z
.....

 

[currysauce]

thankyou good sir

I have a further 2 questions

1. Expand (x - 1/x)^3 in ascending powers of 3.... DID THAT. ii) If x - 1/x = 1, find the value of x^3 - 1/x^3



2.

The lengths of the sides of a scalene triangle are in arithmetic progression. If the largest angle in such a triangle measures 120 degrees, show that the smallest angle measures approx. 22 degrees.

 

[香港!]

thankyou good sir

I have a further 2 questions

1. Expand (x - 1/x)^3 in ascending powers of 3.... DID THAT. ii) If x - 1/x = 1, find the value of x^3 - 1/x^3



2.

The lengths of the sides of a scalene triangle are in arithmetic progression. If the largest angle in such a triangle measures 120 degrees, show that the smallest angle measures approx. 22 degrees.

lol u asked question 1. before!!
i did it and shafqat did another solution as well!

 

[香港!]

http://community.boredofstudies.org/showthread.php?t=82812
there

 

[currysauce]

can aynone get the last 2 questions?

 

[香港!]

"2.

The lengths of the sides of a scalene triangle are in arithmetic progression. If the largest angle in such a triangle measures 120 degrees, show that the smallest angle measures approx. 22 degrees."

let the sides be a-d, a, a+d
draw it....
then we can get by the cosine rule...
cos120=[(a²)+(a-d)²-(a+d)²]\2a(a-d)
=(a²-4ad)\(2a²-2ad)=-1\2
(a-4d)\(a-d)=-1
a-4d=-a+d
2a=5d
d=2a\5
so now the sides of that scalene triangle are
3a\5, a, 7a\5

now using the sine rule...
(7a\5)\sin120=(3a\5)\sin@ where @ is that smallest angle required
7\sin120=3\sin@
sin@=(3sqrt3)\14
@=21'47' by calculator

 

[currysauce]

thankyou...

any luck on the series question right up the first post?

 

[香港!]

thankyou...

any luck on the series question right up the first post?

"In an arithmetic progression, whose first term and common difference are both non-zero, Un denotes the nth term and Sn denotes the sum of n terms. If U6, U4, U10 for a geometric progression:

i) show that S10=0
ii) show that S6 + S12 = 0
iii) deduce that U7 + U8 + U9 + U10 = U11 +U12"


lol..."If U6, U4, U10 for a geometric progression:"
that doesn't make sense... not to me anyway =P
anyway, me doing english now, sum 1 else can help u!!
kekeke

 

[word.]

In an arithmetic progression, whose first term and common difference are both non-zero, Un denotes the nth term and Sn denotes the sum of n terms. If U6, U4, U10 for a geometric progression:

i) show that S10=0
ii) show that S6 + S12 = 0
iii) deduce that U7 + U8 + U9 + U10 = U11 +U12


i)
U₆ = a + 5d
U₄ = a + 3d
U₁₀ = a + 9d

It's in geometric progression so U₄/U₆ = U₁₀/U₄

^{a + 3d}/_{a + 5d} = ^{a + 9d}/_{a + 3d}

a² + 6ad + 9d² = a² + 14ad + 45d²

a = ^-9d/₂

S_n = ⁿ/₂[2a + (n - 1)d]
S₁₀ = 5(-9d + 9d) = 0


ii)
S₆ = 3(-9d + 5d) = -12d
S₁₂ = 6(-9d + 11d) = 12d

hence S₆ + S₁₂ = 0

iii)
U₇ + U₈ + U₉ + U₁₀ = S₁₀ - S₆

U₁₁ + U₁₂ = S₁₂ - S₁₀

From i, S₁₀ = 0
From ii, S₆ + S₁₂ = 0
so S₆ = -S₁₂

Therefore U₇ + U₈ + U₉ + U₁₀ = U₁₁ +U₁₂