Sequence and seriesssss (1 Viewer)

[fullonoob]

Suppose Tn = ar^n-1 is a GP with first term a and ratio r. Show that the sequence Un = log(base2) Tn is an AP, and find its first term and difference.

Help again please.

 

[Drongoski]

Suppose Tn = ar^n-1 is a GP with first term a and ratio r. Show that the sequence Un = log(base2) Tn is an AP, and find its first term and difference.

Help again please.

.: U_n = (n-1)log₂ r + log₂ a

.: U_n+1 - U_n = { (n+1 - 1)log₂ r + log₂} - {(n-1)log₂ r + log₂ a}

=log₂ r (a constant)

= the common difference

.: U_n is an AP.

1st term = log₂ a

 

[Aquawhite]

^Very nice

 

[fullonoob]

Okk thx heres another one
Let triangle ABC be right angled at C, so a^2 + b^2 = c^2. Find the ratio c:a if:
a) b is the AM of a and c
b) b is the GM of a and c
Diagram would help

 

[fullonoob]

bump bump help please

 

[Drongoski]

Okk thx heres another one
Let triangle ABC be right angled at C, so a^2 + b^2 = c^2. Find the ratio c:a if:
a) b is the AM of a and c
b) b is the GM of a and c
Diagram would help

Trying out a) :

b = (a+c)/2

.: a^2 + [(a+c)/2]^2 = c^2

x 4 ==> 4a^2 + (a+c)^2 = 4c^2

.: 3c^2 -5a^2 - 2ac = 0

dividing by a^2: 3(c/a)^2 - 2(c/a) - 5 = 0

This is a quadratic eqn like 3m^2 - 2m - 5 = 0 [ where m = c/a]

Here (3m-5)(m+1) = 0

.: m = 5/3 or m = -1 (-1 inadmissible; ratio of sides of triangle non-neg)

.: m = c/a = 5/3



I'll let somebody else do b)

 

[shaon0]

b^2=ac
a^2+ac=c^2
a^2+ac-c^2=0
(a+[c/2])^2=5c^2/4
a+[c/2]=sqrt(5)c/2
a=(c/2)[sqrt(5)-1]
a[sqrt(5)+1]/2=c
c/a=[sqrt(5)+1]/2

Idk, if this is correct though

 

[Lukybear]

.: U_n = (n-1)log₂ r + log₂ a

.: U_n+1 - U_n = { (n+1 - 1)log₂ r + log₂} - {(n-1)log₂ r + log₂ a}

Is this step neccessary? Can we just state that a = log2a and d = log2r since, its in the form of a+(n-1)d?

Okk thx heres another one
Let triangle ABC be right angled at C, so a^2 + b^2 = c^2. Find the ratio c:a if:
a) b is the AM of a and c
b) b is the GM of a and c
Diagram would help

What does b is the AM of a and c mean?

 

[ninetypercent]

^b is the arithmetic mean of a and c

 

[Lukybear]

ooo i see. I thought it meant arithmetic progression

but how can b be a geometric mean?

 

[Drongoski]

Is this step neccessary? Can we just state that a = log2a and d = log2r since, its in the form of a+(n-1)d?


Interesting. You've got a point.

 

[fullonoob]

Hmmm i'm not sure of another question.
The sequence Tn = 2x3^n + 3x2^n is the sum of two GPs. Find Sn.

This is what i have but it doesnt match the answer.
Sn = 3^(n+1) - 1 + 3 (2^(n+1) - 1)
= 3^n x 3 + 6 x 2^n -4


The answer is Sn = 3^n x 3 + 6 x 2^n -9

I was thinking it was something to do with the n+1 as if the initial numbers (a) affected it. Because a = 2 in the first part of the GP and a = 3 on the second part would make a = 5 o-o
And if i minus 5 from my answer i'd get the right answer xD
Explain please im lost.

 

[fullonoob]

A GP has a positive first term a, and has a limiting sum S(infinite). Show that S(infinite) > (1/2)a

Mehhhhhh proving questions

 

[fullonoob]

find the condition for this Gp to have a limiting sum, then find that limiting sum

1+(1/5x) + (1/(5x)^2) + .....

I just dont get the find the condition part
HELLLLPPP

Spammage questions much xD

 

[Drongoski]

Hmmm i'm not sure of another question.
The sequence Tn = 2x3^n + 3x2^n is the sum of two GPs. Find Sn.

This is what i have but it doesnt match the answer.
Sn = 3^(n+1) - 1 + 3 (2^(n+1) - 1)
= 3^n x 3 + 6 x 2^n -4


The answer is Sn = 3^n x 3 + 6 x 2^n -9

I was thinking it was something to do with the n+1 as if the initial numbers (a) affected it. Because a = 2 in the first part of the GP and a = 3 on the second part would make a = 5 o-o
And if i minus 5 from my answer i'd get the right answer xD
Explain please im lost.

Very easy!

T_n is made up of 2 GPs

i) with 1st term = 2x3 = 6, r = 3

ii) 1st term = 6, r = 2

.: S_n = A_n + B_n (i.e. sum of the 2 GPs)

= 6(3^n - 1)/(3 - 1) + 6(2^n - 1)/(2 - 1)

= 3(3^n-1) + 6(2^n - 1)

= 3.3^n - 3 + 6.2^n - 6

= 3.3^n + 6.2^n - 9

 

[fullonoob]

Very easy!

T_n is made up of 2 GPs

i) with 1st term = 2x3 = 6, r = 3

ii) 1st term = 6, r = 2

.: S_n = A_n + B_n (i.e. sum of the 2 GPs)

= 6(3^n - 1)/(3 - 1) + 6(2^n - 1)/(2 - 1)

= 3(3^n-1) + 6(2^n - 1)

= 3.3^n - 3 + 6.2^n - 6

= 3.3^n + 6.2^n - 9

ahahaha yeah sorry. Got it after a while but didnt remove post. How about the other one, finding conditions