Alpha Sequences and Series—SOLNS
_ FAMAT State Convention 2017
1. A—There are 9 numbers with an arithmetic seq. of difference 0 (1111 through 9999). There are
6 with a difference 1 (1234 through 6789). There are 3 with a difference 2 (1357 through 3579).
There are 7 with a difference −1 (3210 through 9876). There are 4 with a difference −2 (6420
through 9753), and there is 1 with a difference of −3 (9630). The answer is therefore 9 + 6 + 3
+ 7 + 4 + 1 =30.
a
24

 72. Doubling that for the
1 r
1  (2 / 3)
return bounces back up and then adding on the original drop of 36 gives the answer of 180.
2. C— The sum of the bounces is given by the formula S 
3. D—Suppose that facing the first flight of stairs is North. Then every 4 floors brings you 2ft West
and 2ft South, by the end of 100 floors, you are
100
 i  50 x 101 ft. high.
The total distance is
i 1
then
502  502  (50 x 101) 2  50 1  1  10201  50 10203.
4. E—(4) The first few terms of the sequence are 10, 5, 12, 6, 3, 8, 4, 2, 1, 4, 2, 1, … Since the
sequence repeats every three terms (after the 6th term) and 2008 is one more than a multiple of
3, the 2008th term has the same value as the 7th term.
5. B—If S ( n ) is the nth partial sum, note that if m is the k th triangular number, then S (m)  k 2 .
44(44  1)
Since 442  1936 and 452  2025, we want to begin our search at
= 990. Because
2
(2010  1936)
 37, 37 more 2s are needed, so the needed term is n = 990 + 37 = 1027.
2
 1 (1  3  5...  99)   1 502   1 2500 
 1 a  1 b   1 (a  b) 
6.C—Note that 



 . So, 


1 
1
1 
 0 1  0 1   0
0
 0 1  0
7
7. C— 400  (101  a7 )  a7  156.14.
2
8. C—Using the recursive formula and the given initial values, we can find that a4  7, a5  9
and an  2n  1. So, we are looking for the sum of the 1st 30 odd integers, which is 302  900.
9. A—Rationalizing denominators,

 
2 1 
 
3 2 

4  3  ... 


25  24  25  1  4.
Page 1 of 4
10. B—We need to minimize 24a 15b or 3 8a  5b , for positive integers a and b. So this is
always a multiple of 3, and thus no smaller than 3.
11. B—There are 2 ways to attack this problem, one with using the arithmetic sum formulas
and the other way is to realize that every integer in the second sum is n more than the
corresponding integer in the first sum. Since there are n integers in the first sum, we get:
Sum1  100  Sum1  n 2  n  10.
12. C—We have  a1   49, so a2  4  9  13   a2   169, so a3  16   a3   256, so a4  13, and
2
2
2
so on. Since ai depends only on ai 1 , the sequence is 7, 13, 16, 13, 16, 13, 16,… and a2015 = 16.
13. D—If the roots are r  d , r , and r  d , then the sum of the roots is 3r  6, so r  2. Substitution
this into the original equation for x gives c  64.
14. D—The first few triangular numbers are 1,3, 6,10,15, 21, 28,36, 45,... It is easily seen that 2 out
of 3 every triangular numbers are multiples of 3. These are of the form 3k – 1 or 3k. Since
2
100 is not of this form, there are (99)  66 multiples of 3 in the described set.
3
15. D—Using sin x  cos(90  x) , we get cos2 (80 )  ...  cos2 (50 )  sin 2 (50 )  ...  sin 2 (90 )  5
16. B—49 dragons  49(7) = 343 sheep killed  343(7) = 3401 ears of corn would have been
eaten  2401(7) = 16807 lbs. of grain.
17. A—Finishing the list of terms for the geometric series, we get: 1, 2, 4, 8, 16, 32. Their sum
6
is 63. Using the arithmetic sum formula, we get 63  (1  x)  x  20.
2
18. C—This is a geometric series and for convergence, the sum must be S 
a
, with S =2,
1 r
2
a  x, and r  x. Plugging in, this yields x  .
3
19. B—There are several ways to break this into 2 series, but I chose:
1
2 2 2
 1 1
 18 3 21

 ...     .
    ...    
 1 9 81
  3 27 243
 8 8 8
Page 2 of 4
20. A—The reciprocals of this sequence form an arith. prog. with the 1st term and the common
1
3
1
1 4
diff. both = . The 8th term is  7    . So, the 8th term in the harmonic series is .
6
4
6
6 3
21. D—We have 1  2i  3i 2  ...  (n  1)i n  [1  2i  3  4i]  [5  6i  7  8i]  ...  (n  1), where there are
n
of the expressions in brackets and an isolated ( n  1). Each bracketed expression adds to
4
n(2i  2)
(n  2  ni )
 (n  1) 
.
( 2i  2), so the sum is
4
2
2
 2  3  4   n  1 
22. B—The product is     ... 
 , which telescopes to .
n
 3  4  5   n 
a
a
, a, ar . So,  a  ar  a 3  1  a  1. Also,
r
r
a
1 7
1
 a  ar  r  1    r  2 or . So, the roots are 2, 1, ½ and the sum is 3.5.
r
r 2
2
23. C—Let the 3 roots be
(a  c)
. Also, since a  1, b, c and a, b, c  2
2
are geometric sequences, we know b  (a  1)c  a(c  2). Squaring,
24. B—Since a, b, c is an arithmetic sequence, then b 
a 2  2ac  c 2
 ac  c  ac  2a. This gives c  2a. Subbing this into the 1st equality, gives
4
ac
a  8, so c  16 and b 
 12.
2
25. A—The sum of the first 3 terms is a  ar  ar 2 which = 7a  1  r  r 2  7  r  2, 3. Ignore
the negative case (it says positive). We also know that a  ar  ar 2  ar 3  45. Sub in
r  2 gives a  3.
26. B—The nth angle of the polygon is 160  5(n  1)  165  5n and the sum of the angle measures
n
is (160  165  5n) which equals ( n  2)180. Solving that equality gives n  9.
2
1 1
1
1


 1  4
27. B—The sum, S  1           ...   1  2  4  ...    
.

2
4
 2 2

 1  1/ 4  3
2
2
2
28. B—If the numbers are x and y, we have
x y
 xy  50. So, x  y  2 xy  100 or
2
( x  y )2  100  x  y  10.
Page 3 of 4




 1 1
 1 1  1 1   1   1  1  15
29. B—The required sum is 1    ... 1   ... 1   ...   

 .

 2 4
 3 9  5 25   1  1   1  1  1  1  4
 2  3  5 
30. C—Sum =
n
1000
(a  l ) 
(2  2000)  1001000 .
2
2
Page 4 of 4