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134
Solutions to Exercises
of
k
+ 1 terms equals the sum of
k
terms plus the
(k
+ l)st term. The induction hypothesis, together
with the fact that
ak+1
=
a
+
kd
says that this is equal to
k
(k(k1) )
2[2a
+
(k
l)d)
+
a
+
kd
=
ak
+
a
+
2
+
k d
=ak+a+
(k(k;l))d=
(k;1)(2a+kd)
as required. By the Principle of Mathematical Induction, the result is true for all
n
~
1.
23. Let the sequence be
a, a
+
d, a
+
2d,
. .
.. We are given 1
2
0
(2a
+
9d)
=
2
~
(2a
+
5d),
that is,
20
lOa
+
45d
=
12a
+
30d,
or
2a
=
15d.
Also,
"2(2a
+
19d)
= 1360, so
20a
+
190d
= 1360.
Substituting
20a
=
10(2a)
=
150d,
we obtain
340d
= 1360, so
d
= 4 and
a
=
125d
=
30. The sum
of
n
terms of the sequence is
~[60
+
4(n

1))
=
~(4n
+ 56)
=
2n
2
+
28n.
24. Let
x, y, z, w
be
a, a+d, a+ 2d, a+3d,
respectively. We are given
that4a+6d
=
8, so
d
=
~(2
a).
We are also given that
a( a
+
3d)
+
(a
+
d) (a
+
2d)
= 
2, which is
a
2
+
3ad
+
d
2
=
1. Substituting
d
=
~(2
a),
we obtain
a
2
+2(2
a)a+
~(4
4a+a
2
)
=
1, so
a
2
4a
5
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 Summer '10
 any
 Graph Theory

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