Please point me in the right direction!

Maryanne

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This question seems impossible, if someone just starts me off I know i'll eventually get it!

Determine the geometric sequence with each property:

a) the sum of the 3rd and 4th terms is 36. The sum of the 4th and 5th terms is 108.

b) The sum of the first 2 terms is 3. The Sum of the next 2 terms is 4/3

Thanks!
 
The terms are A,B & C
A+B=36
B+C=108
B=x*A
C=x*B
Four equations in four unknowns. Substitution seems indicated?

Try b) with that as a model.
 
Or use the formulas you've learned. For instance:

. . . . .a<sub>n</sub> = a<sub>n-1</sub>r

Then:

. . . . .a<sub>3</sub> + a<sub>4</sub> = a<sub>4</sub>/r + a<sub>4</sub> = 36

. . . . .a<sub>4</sub>(1/r + 1) = 36

. . . . .a<sub>4</sub> = (36r)/(1 + r)

Now follow a similar procedure with the fourth and fifth terms, finding another expression for a<sub>4</sub> in terms of r. Solve for r. Back-solve for the various values you need.

Eliz.
 
Looks to me like Eliz and I said exactly the same thing. You can't go wrong now. x=r
----------------
Gene
 
O.k so if I use t (to represent term)

t3+t4= 36

therefore t3+rt3=36

t4+t5=108

t3, rt3, r²t3...........

what formula would I then use?

I know the sequence is 1, 3, 9, 27, 81........
 
This question seems impossible, if someone just starts me off I know i'll eventually get it!
Determine the geometric sequence with each property:
a) the sum of the 3rd and 4th terms is 36. The sum of the 4th and 5th terms is 108.


Just another viewpoint:

1--Let the sequence be A, B, C, D, E,...
2--Then, C + D = 36 and
3--D + E = 108
4--Let the common ratio be k
5--Then, c + ck = 36 or c(1 + k) = 36 and
6--d + dk = 108 or d(1 + k) = 108
7--These yield 36/c = 108/d or 108c = 36d or d = 3c
8--With k = 3, c + 3c = 36 or 4c = 36 making c = 9, d = 27 and e = 81.
9--9 + 27 = 36 and 27 + 81 = 108
10--The geometric series is 1, 3, 9, 27, 81, 243, ....
 
You are right.
The very simplest way is to note
t<sub>3</sub> + t<sub>4</sub> = 36
t<sub>4</sub> = r*t<sub>3</sub> so
t<sub>3</sub> + r*t<sub>3</sub> = 36
t<sub>4</sub> + t<sub>5</sub> = 108
t<sub>4</sub> + r*t<sub>4</sub> =
r*(t<sub>3</sub> + rt<sub>4</sub>) =
r*36 = 108
r=3
The general term is t<sub>n</sub> = r^(n-1)*t<sub>1</sub>
 
This question seems impossible, if someone just starts me off I know i'll eventually get it!
Determine the geometric sequence with each property:
b) The sum of the first 2 terms is 3. The Sum of the next 2 terms is 4/3

1--Let the sequence be A, B, C, D
2--Then, A + B = 3 = 9/3 and
3--C + D = 4/3
4--A + kA = 9/3 and C + kC = 4/3
5--From these comes 4A = 9C or C = 4A/9 and B = 2A/3
6--A + 2A/3 = 9/3 yielding A = 9/5
7--It follows that B = 18/15, C = 36/45 and D = 72/135
8--9/5 + 18/15 = 45/15 = 3 and 36/45 + 72/135 = 180/135 = 4/3.
 
Just to stir the pot, carrying my method one step further gives
r^2*3=4/3
r=+2/3
A=9 & r=-2/3
gives a second progression
9,-6,4,-8/3,...
which also satisfies the problem unless there is a law I've forgotten that r must be positive :evil:
----------------
Gene
 
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