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> | + | 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>. |
== See Also == | == See Also == |
Revision as of 01:51, 1 June 2021
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
.
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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