Difference between revisions of "1969 AHSME Problems/Problem 33"
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== Solution 2 (Quick Sol) == | == Solution 2 (Quick Sol) == | ||
Note that the sum of arithmetic sequences can be expressed as a quadratic polynomial of <math>n</math>. From the ratio given in the problem, multiply <math>n</math> to both sides to get quadratic polynomials. <cmath>S_n = 7n^2+n, T_n = 4n^2+27n</cmath> From there, the 11th term for <math>S_n</math> and <math>T_n</math> can becalculated. <cmath>S_{11} - S_{10} = 7*11^2+11 - (7*10^2+10) = 148</cmath> <cmath>T_{11} - T_{10} = 4*11^2+27*11 - (4*10^2+27*10) = 111</cmath> The ratio is <math>148 : 111 = \boxed{\textbf{(A) } 4:3}</math>. | Note that the sum of arithmetic sequences can be expressed as a quadratic polynomial of <math>n</math>. From the ratio given in the problem, multiply <math>n</math> to both sides to get quadratic polynomials. <cmath>S_n = 7n^2+n, T_n = 4n^2+27n</cmath> From there, the 11th term for <math>S_n</math> and <math>T_n</math> can becalculated. <cmath>S_{11} - S_{10} = 7*11^2+11 - (7*10^2+10) = 148</cmath> <cmath>T_{11} - T_{10} = 4*11^2+27*11 - (4*10^2+27*10) = 111</cmath> The ratio is <math>148 : 111 = \boxed{\textbf{(A) } 4:3}</math>. | ||
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+ | == Solution 3 (Quickest Sol) == | ||
+ | Since they are arithmetic sequences, we know that the 11th term is the average value of the first 21 terms in each sequence. So the desired ratio of the 11th terms is just <math>(21 \cdot 7 + 1) : (21 \cdot 4 + 27) = 148 : 111 = \boxed{\textbf{(A) } 4:3}</math> | ||
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+ | -purplepenguin2 | ||
== See Also == | == See Also == |
Latest revision as of 23:44, 3 April 2023
Problem
Let and
be the respective sums of the first
terms of two arithmetic series. If
for all
, the ratio of the eleventh term of the first series to the eleventh term of the second series is:
Solution
Let be the first arithmetic sequence and
be the second arithmetic sequence. If
, then
. Since
and
are just the first term, the first term of
is
and the first term of
is
for some
. If
, then
, so the sum of the first two terms of
is
and the sum of the first two terms of
is
for some
. Thus, the second term of
is
and the second term of
is
, so the common difference of
is
and the common difference of
is
.
Thus, using the first terms and common differences, the sum of the first three terms of equals
, and the sum of the first three terms of
equals
. That means
With the substitution, the common difference of
is
, and the common difference of
is
. That means the
term of
is
, and the
term of
is
. Thus, the ratio of the eleventh term of the first series to the eleventh term of the second series is
.
Solution 2 (Quick Sol)
Note that the sum of arithmetic sequences can be expressed as a quadratic polynomial of . From the ratio given in the problem, multiply
to both sides to get quadratic polynomials.
From there, the 11th term for
and
can becalculated.
The ratio is
.
Solution 3 (Quickest Sol)
Since they are arithmetic sequences, we know that the 11th term is the average value of the first 21 terms in each sequence. So the desired ratio of the 11th terms is just
-purplepenguin2
See Also
1969 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 32 |
Followed by Problem 34 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
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