IB HL Math Homework #1 Solutions
Assigned: 8/21/07, Tuesday
Due: 8/24/05, Friday
1) (i) A rubber ball is dropped from a height of 81 meters. Each time it strikes the ground
it rebounds two thirds of the distance through which it has fallen.
(a) Find the maximum height of the ball between the fifth bounce and the sixth
bounce.
(b) What is the total distance traveled by the ball from when it is dropped to the
time it strikes the ground for the sixth time?
(c) Assume that the ball continues to bounce indefinitely. What is the total
distance traveled by the ball?
Solution
(a) Define a sequence a1, a2, etc. such that ai equals the maximum height of the ball
in between bounce number i and bounce number i+1. This sequence would be a
geometric one, because each bounce reduces the height of the ball by a constant
ratio. To solve this part, we must ascertain the value of a5. We note that a1 = 54 m,
which is two-thirds of the initial height of 81 m. We also note that r = 2/3. Now we
can solve for a5:
2
32
a5  a1 r 4  54( ) 4 
m
3
3
(b) After the initial 81 m, each subsequent trip of the ball involves the ball bouncing
up and then coming back down. Thus, the sum we desire is as follows:
54(1  ( 2 ) 5
1087
3 )  81  6(54)(1  32
81   2ai  81  2 ai  81  2(
)
m
243
2
3
1
i 1
i 1
3
(c) We can use the framework for our solution for b, but just set the sum part of it
up to go to infinity:


54(1)
81   2ai  81  2 ai  81  2(
)  81  6(54)  405m
1 2
i 1
i 1
3
5
5
This works because by setting the sum to go to infinity, we are adding up the sum of
the distances the ball travels up and down, after the initial fall of 81 meters. In real
life of course, the ball would stop and the actual distance traveled would have an
upper bound of 405 meters.
(ii) The sum of the first three numbers in an arithmetic sequence is 24. If the first
number is decreased by 1 and the second number is decreased by 2, then the third number
and the two new numbers are in geometric sequence. Find all possible sets of three
numbers which are in the arithmetic sequence.
Solution
Let the numbers in the arithmetic sequence by a, b and c, and let their common
difference be d. It follows that a = b – d, and c = b + d.
Thus, a + b + c = b – d + b + b + d = 3b = 24, so b = 8.
Furthermore, we are given that a – 1, b – 2 = 6, and c are in geometric sequence, so
we get the equations:
6
c

a 1 6
Using that a – 1 = 8 – d – 1 = 7 – d, and that c = 8 + d, we have:
6
8d

7d
6
36  (7  d )(8  d )
36  56  d  d 2
d 2  d  20  0
(d  4)( d  5)  0
So d = 4 or d = -5. The first leads to the terms 4, 8, 12 and the second value of d leads
to the terms 13, 8, 3.
2) (2002 IB HL P1 #1) Consider the arithmetic series 2 + 5 + 8 + ….
a) Find an expression for Sn, the sum of the first n terms.
b) Find the value of n for which Sn = 1365.
Solution
Let the first term of the sequence be a1 and the common difference be d. In general,
we find that an = a1 + (n-1)d = 2 + (n-1)3 = 3n – 1. Thus, we can solve for Sn:
Sn 
(a1  an )n n(2  3n  1) n(3n  1)
.


2
2
2
Now, we find the value of n for which Sn = 1365:
n(3n  1)
 1365
2
3n 2  n  2730  0
(3n  91)( n  30)  0 , since n>0, n = 30.
Sn 
3) (1999 IB HL P2 #2i) The ratio of the fifth term to the twelfth term of a sequence in an
arithmetic progression is 6/13. If each term of this sequence is positive, and the product
of the first term and the third term is 32, find the sum of the first 100 terms of this
sequence.
Solution
Let the arithmetic progression have the first term a1 and a common difference of d.
We can set up the two following equations:
a1  4d
6

a1  11d 13
13a1  52d  6a1  66d
7a1  14d
a1  2d , substitute this value into
the second equation.
S100 
(a1  a100 )100
 50(4  202)  10300
2
a1 (a1  2d )  32
2d (2d  2d )  32
8d 2  32
d  2 , since d > 0.
and a1  4 .
4) (1993 AHSME #21) Let a1, a2, … , ak be a finite arithmetic sequence with
14
a4 + a7 + a10 = 17 and
a
i4
i
 77 .
If ak = 13, what is the value of k?
Solution
Let d be the common difference of the sequence and substitute into the given
equations:
14
a4  a7  a10  17
a
a1  3d  a1  6d  a1  9d  17
a
3a1  18d  17
11a1  88d  77
18(11)
17(11)
 11a1 
d 
3
3
----------------------------------44
2
22d 
, so d 
3
3
i
 77
1
 (i  1)d  77
i4
14
i4
2
17  18( )
3  5 . Now, solve for k:
Then we have that a1 
3
3
ak 
5
2
 (k  1)  13 , so k  18 .
3
3