# Math Help - Geometric Sequences 1

1. ## Geometric Sequences 1

A ball is dropped from a height of 3 m. After each bounce it rises to 75% of its previous height.

a) State t1 and r: t1 = 3 m r = 0.75

b) Write the general term for the sequence:
tn = (t1)(r^n-1)
tn = (3)(0.75^n-1)

c) What height does the ball reach after the 6th bounce? (I am not sure how to do this question)
This was my attempt: I suppose you solve for tn and you plug in 6 into n...
tn = (3)(0.75^6-1)
tn = (3)(0.75^5)
tn = (3)(.2373046875)
tn = .7119140625

My answer was wrong, the answer is 53.39 cm.

d) After how many bounces will the ball reach a height of approx. 40 cm?
(I cannot solve this one either)
My attempt: I plugged in 40 cm into tn and try to solve for n:
40 = (3)(0.75^n-1)
13.3 = 0.75^n-1 I am stuck here... I do not know how to get n alone with algebra....

The answer is 7 times.

Could anyone tell me where I went wrong for c) and d) and what I should have done? Thanks!

2. ## Re: Geometric Sequences 1

I'm not sure why you put "n - 1" in the general term. To get your books answer, just multiply 3/4 six times.

On the second part ... you need a logarithmic calculation. See the attached pdf.

3. ## Re: Geometric Sequences 1

Originally Posted by Mathnood768
A ball is dropped from a height of 3 m. After each bounce it rises to 75% of its previous height.

a) State t1 and r: t1 = 3 m r = 0.75

b) Write the general term for the sequence:
tn = (t1)(r^n-1)
You are numbering the bounces incorrectly. With tn= (t1)(r^(n-1)) n= 1 initially. After the first bound n= 2, after the third bounce, n= 4, etc.
Another way to handle this is to take the initial height as t0 so that t1 is after the first bounce, etc. Then you can use tn= (t1)(r^n).

tn = (3)(0.75^n-1)

c) What height does the ball reach after the 6th bounce? (I am not sure how to do this question)
This was my attempt: I suppose you solve for tn and you plug in 6 into n...
tn = (3)(0.75^6-1)
tn = (3)(0.75^5)
tn = (3)(.2373046875)
tn = .7119140625
After the sixth bounce n= 7 so t7= (3)(0.75^(7-1))= (3)(0.75^6)

My answer was wrong, the answer is 53.39 cm.

d) After how many bounces will the ball reach a height of approx. 40 cm?
(I cannot solve this one either)
My attempt: I plugged in 40 cm into tn and try to solve for n:
40 = (3)(0.75^n-1)
13.3 = 0.75^n-1 I am stuck here... I do not know how to get n alone with algebra....
40 cm is .4 meters. Your equation should be .4= (3)(0.75^(n-1). Then .4/3= 0.133333= (0.75)^(n-1). Since the "n" is an exponent, take the logarithm of both sides. log(0.133333)= (n- 1) log(0.75) so n= 1+ log(0.133333)/log(0.75). But remember that this will be after the n- 1 bounce.

The answer is 7 times.

Could anyone tell me where I went wrong for c) and d) and what I should have done? Thanks!

4. ## Re: Geometric Sequences 1

Thanks! This makes sense to me, and I have one more question,

40 cm is .4 meters. Your equation should be .4= (3)(0.75^(n-1). Then .4/3= 0.133333= (0.75)^(n-1). Since the "n" is an exponent, take the logarithm of both sides. log(0.133333)= (n- 1) log(0.75) so n= 1+ log(0.133333)/log(0.75). But remember that this will be after the n- 1 bounce.

So, I got 8 in the calculator, should I just minus 1 for the final answer? Am I allowed to do that? And I should do that every time I have a question like this right?