Difference between revisions of "1969 AHSME Problems/Problem 33"

Latest revision as of 22:44, 3 April 2023

Problem

Let $S_n$ and $T_n$ be the respective sums of the first $n$ terms of two arithmetic series. If $S_n:T_n=(7n+1):(4n+27)$ for all $n$ , the ratio of the eleventh term of the first series to the eleventh term of the second series is:

$\text{(A) } 4:3\quad \text{(B) } 3:2\quad \text{(C) } 7:4\quad \text{(D) } 78:71\quad \text{(E) undetermined}$

Solution

Let $S$ be the first arithmetic sequence and $T$ be the second arithmetic sequence. If $n = 1$ , then $S_1:T_1 = 8:31$ . Since $S_1$ and $T_1$ are just the first term, the first term of $S$ is $8a$ and the first term of $T$ is $31a$ for some $a$ . If $n = 2$ , then $S_2:T_2 = 15:35 = 3:7$ , so the sum of the first two terms of $S$ is $3b$ and the sum of the first two terms of $T$ is $7b$ for some $b$ . Thus, the second term of $S$ is $3b-8a$ and the second term of $T$ is $7b - 31a$ , so the common difference of $S$ is $3b-16a$ and the common difference of $T$ is $7b-62a$ .

Thus, using the first terms and common differences, the sum of the first three terms of $S$ equals $\tfrac{1}{2} \cdot 3(16a + 2(-16a + 3b))$ , and the sum of the first three terms of $T$ equals $\tfrac{1}{2} \cdot 3(62a + 2(-62a + 7b))$ . That means $\frac{16a + 2(-16a + 3b)}{62a + 2(-62a + 7b)} = \frac{22}{39}$ $\frac{6b-16a}{14b-62a} = \frac{22}{39}$ $\frac{3b-8a}{7b-31a} = \frac{22}{39}$ $117b - 312a = 154b - 682a$ $-37b = -370a$ $b = 10a$ With the substitution, the common difference of $S$ is $14a$ , and the common difference of $T$ is $8a$ . That means the $11^\text{th}$ term of $S$ is $8a + 10(14a) = 148a$ , and the $11^\text{th}$ term of $T$ is $31a + 10(8a) = 111a$ . Thus, the ratio of the eleventh term of the first series to the eleventh term of the second series is $148a:111a = \boxed{\textbf{(A) } 4:3}$ .

Solution 2 (Quick Sol)

Note that the sum of arithmetic sequences can be expressed as a quadratic polynomial of $n$ . From the ratio given in the problem, multiply $n$ to both sides to get quadratic polynomials. $S_n = 7n^2+n, T_n = 4n^2+27n$ From there, the 11th term for $S_n$ and $T_n$ can becalculated. $S_{11} - S_{10} = 7*11^2+11 - (7*10^2+10) = 148$ $T_{11} - T_{10} = 4*11^2+27*11 - (4*10^2+27*10) = 111$ The ratio is $148 : 111 = \boxed{\textbf{(A) } 4:3}$ .

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 $(21 \cdot 7 + 1) : (21 \cdot 4 + 27) = 148 : 111 = \boxed{\textbf{(A) } 4:3}$

-purplepenguin2

@@ Line 23: / Line 23: @@
 == 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>.
+== 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>
+-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
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