Ma 310 Exam 2 Worksheet With Answers - University Of Kentucky Page 4

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One way to do this is to see from the diagrams that the numbers of toothpicks follow
the pattern:
Pattern Toothpicks
1 3
2 3 + 6
3 3 + 6 + 9
4 3 + 6 + 9 + 12
So Pattern #100 will have 3 + 6 + 9 + 12 + · · · + 300 toothpicks. This equals 3(1 + 2 +
100×101
3 + · · · + 100) = 3
= 15150.
2
8. Suppose a
, a
, a
, . . . is a geometric sequence, a
= 3a
, and a
= 10. Find the exact
0
1
2
9
7
3
value of a
(do not express this as an approximation). Show your work.
8
2
Because this is a geometric sequence we have a
= a, a
= ar, a
= ar
, etc. In general
0
1
2
n
9
7
7
2
1/2
a
= ar
. Now a
= 3a
so ar
= 3ar
. Dividing by ar
, r
= 3 so r = 3
. Using
n
9
7
3
3/2
−3/2
8
= 10, so a × 3
= 10, a = 10 × 3
a
= 10, we have ar
. Finally a
= ar
=
3
8
−3/2
8/2
5/2
1/2
10 × 3
× 3
= 10 × 3
= 90 × 3
.
9. You have a square-based pyramid. The base is a square of side length 15. From the
top of this large pyramid you cut off a smaller square-based pyramid whose base is a
square of side length 5. The height of the remaining solid is 12. What is the volume of
1
the remaining solid? Show your work. Note: The volume of a pyramid is
Bh, where
3
B is the area of the base and h is the height.
Below is a diagram of the cross-section. AB = 15, CD = 5, and F G = 12.
E
G
D
C
A
B
F
x
x+12
Let x = EG. Using similar triangles ΔCDE ∼ ΔABE we have
=
. Solving,
5
15
15x = 5x + 60 so x = 6. Now subtract the volume of the small pyramid from the
1
1
2
2
(18) −
volume of the large pyramid:
15
5
(6) = 1300 cubic units.
3
3

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