Difference between revisions of "1984 AIME Problems/Problem 1"
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Find the value of <math>a_2+a_4+a_6+a_8+\ldots+a_{98}</math> if <math>a_1</math>, <math>a_2</math>, <math>a_3\ldots</math> is an [[arithmetic progression]] with common difference 1, and <math>a_1+a_2+a_3+\ldots+a_{98}=137</math>. | Find the value of <math>a_2+a_4+a_6+a_8+\ldots+a_{98}</math> if <math>a_1</math>, <math>a_2</math>, <math>a_3\ldots</math> is an [[arithmetic progression]] with common difference 1, and <math>a_1+a_2+a_3+\ldots+a_{98}=137</math>. | ||
− | + | == Solution 1 == | |
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One approach to this problem is to apply the formula for the sum of an [[arithmetic series]] in order to find the value of <math>a_1</math>, then use that to calculate <math>a_2</math> and sum another arithmetic series to get our answer. | One approach to this problem is to apply the formula for the sum of an [[arithmetic series]] in order to find the value of <math>a_1</math>, then use that to calculate <math>a_2</math> and sum another arithmetic series to get our answer. | ||
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A somewhat quicker method is to do the following: for each <math>n \geq 1</math>, we have <math>a_{2n - 1} = a_{2n} - 1</math>. We can substitute this into our given equation to get <math>(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137</math>. The left-hand side of this equation is simply <math>2(a_2 + a_4 + \ldots + a_{98}) - 49</math>, so our desired value is <math>\frac{137 + 49}{2} = \boxed{093}</math>. | A somewhat quicker method is to do the following: for each <math>n \geq 1</math>, we have <math>a_{2n - 1} = a_{2n} - 1</math>. We can substitute this into our given equation to get <math>(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137</math>. The left-hand side of this equation is simply <math>2(a_2 + a_4 + \ldots + a_{98}) - 49</math>, so our desired value is <math>\frac{137 + 49}{2} = \boxed{093}</math>. | ||
− | + | == Solution 2 == | |
If <math> a_1 </math> is the first term, then <math> a_1+a_2+a_3 + \cdots + a_{98} = 137 </math> can be rewritten as: | If <math> a_1 </math> is the first term, then <math> a_1+a_2+a_3 + \cdots + a_{98} = 137 </math> can be rewritten as: | ||
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<math> \frac{137 - 97(49) + 2(49)^2}{2} = \fbox{093} </math>. | <math> \frac{137 - 97(49) + 2(49)^2}{2} = \fbox{093} </math>. | ||
− | + | == Solution 3 == | |
A better approach to this problem is to notice that from <math>a_{1}+a_{2}+\cdots a_{98}=137</math> that each element with an odd subscript is 1 from each element with an even subscript. Thus, we note that the sum of the odd elements must be <math>\frac{137-49}{2}</math>. Thus, if we want to find the sum of all of the even elements we simply add <math>49</math> common differences to this giving us <math>\frac{137-49}{2}+49=\fbox{093}</math>. | A better approach to this problem is to notice that from <math>a_{1}+a_{2}+\cdots a_{98}=137</math> that each element with an odd subscript is 1 from each element with an even subscript. Thus, we note that the sum of the odd elements must be <math>\frac{137-49}{2}</math>. Thus, if we want to find the sum of all of the even elements we simply add <math>49</math> common differences to this giving us <math>\frac{137-49}{2}+49=\fbox{093}</math>. | ||
Or, since the sum of the odd elements if 44, then the sum of the even terms must be <math>\fbox{093}</math>. | Or, since the sum of the odd elements if 44, then the sum of the even terms must be <math>\fbox{093}</math>. | ||
− | + | == Solution 4 == | |
We want to find the value of <math>a_2+a_4+a_6+a_8+\ldots+a_{98}</math>, which can be rewritten as <math>a_1+1+a_2+2+a_3+\ldots+a_{49}+49 \implies a_1+a_2+a_3+\ldots+a_{49}+\frac{49 \cdot 50}{2}</math>. | We want to find the value of <math>a_2+a_4+a_6+a_8+\ldots+a_{98}</math>, which can be rewritten as <math>a_1+1+a_2+2+a_3+\ldots+a_{49}+49 \implies a_1+a_2+a_3+\ldots+a_{49}+\frac{49 \cdot 50}{2}</math>. | ||
We can split <math>a_1+a_2+a_3+\ldots+a_{98}</math> into two parts: | We can split <math>a_1+a_2+a_3+\ldots+a_{98}</math> into two parts: |
Revision as of 04:19, 21 July 2022
Problem
Find the value of if , , is an arithmetic progression with common difference 1, and .
Solution 1
One approach to this problem is to apply the formula for the sum of an arithmetic series in order to find the value of , then use that to calculate and sum another arithmetic series to get our answer.
A somewhat quicker method is to do the following: for each , we have . We can substitute this into our given equation to get . The left-hand side of this equation is simply , so our desired value is .
Solution 2
If is the first term, then can be rewritten as:
Our desired value is so this is:
which is . So, from the first equation, we know . So, the final answer is:
.
Solution 3
A better approach to this problem is to notice that from that each element with an odd subscript is 1 from each element with an even subscript. Thus, we note that the sum of the odd elements must be . Thus, if we want to find the sum of all of the even elements we simply add common differences to this giving us .
Or, since the sum of the odd elements if 44, then the sum of the even terms must be .
Solution 4
We want to find the value of , which can be rewritten as . We can split into two parts: and Note that each term in the second expression is greater than the corresponding term, so, letting the first equation be equal to , we get . Calculating by sheer multiplication is not difficult, but you can also do . We want to find the value of . Since , we find . .
- PhunsukhWangdu
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |