MTH 119 — Spring — 2005
Essex County College — Division of Mathematics
Quiz # 11— May 9, 2005
Name:
Signature:
Show all work clearly and in order, and box your final answers. Justify your answers algebraically
whenever possible. You have 20 minutes to take this 10 point quiz. When you do use your
calculator, sketch all relevant graphs and write down all relevant mathematics.
1. The 3rd term of an arithmetic sequence is 8, and the 16th term is 47. Find a1and d.
Solution: We know the following2
8 = a1+ 2d(1)
47 = a1+ 15d(2)
Which is just two equations and two unknowns, subtracting the first equation from the
second equation will give 39 = 13d, so d= 3. Now substitute d= 3 into any one of these
equations and solve for a1,e.g. using the first equation, 8 = a1+ 6, results in a1= 2. You
should, of course. check to make sure that the solution works for the second equation as
well, i.e. 47 = 2 + 15 ·3. Final answer is d= 3 and a1= 2 .
2. Find the sum of the first 15 terms of the arithmetic sequence
4,7,10,13, . . .
Solution: By inspection, we know that a1= 4 and d= 3, so
S15 = 15 ·4 + (15 −1) (15)
2
·3 = 60 + 315 = 375
So the sum of the first 15 terms is 315 .
3. Find the sum (I’ll accept a correct decimal answer.) of the first 10 terms of the geometric
sequence
16,−8,4,−2, . . .
Solution: By inspection, we know that a1= 16 and r=−
1
2, so
S10 = 16 ·
1−−
1
210
1−−
1
2=341
32
So the sum of the first 10 terms is 341
32 = 10.65625 .
1This document was prepared by Ron Bannon using L
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2an=a1+ (n−1) ·d
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