Difference between revisions of "1984 AIME Problems/Problem 1"
(solution) |
(→Solution 3) |
||
(10 intermediate revisions by 8 users not shown) | |||
Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
+ | |||
+ | === 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 <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. | ||
− | 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} = 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: | ||
+ | |||
+ | <math> 98a_1 + 1+2+3+ \cdots + 97 = 137 </math> <math>\Leftrightarrow</math> | ||
+ | <math> 98a_1 + \frac{97 \cdot 98}{2} = 137 </math> | ||
+ | |||
+ | Our desired value is <math> a_2+a_4+a_6+ \cdots + a_{98} </math> so this is: | ||
+ | |||
+ | <math> 49a_1 + 1+3+5+ \cdots + 97 </math> | ||
+ | |||
+ | which is <math> 49a_1+ 49^2 </math>. So, from the first equation, we know <math> 49a_1 = \frac{137}{2} - \frac{97 \cdot 49}{2} </math>. So, the final answer is: | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | Or, since the sum of the odd elements if 44, then the sum of the even terms must be <math>\fbox{093}</math>. | ||
== See also == | == See also == | ||
− | * [[ | + | {{AIME box|year=1984|before=First Question|num-a=2}} |
− | * [[ | + | * [[AIME Problems and Solutions]] |
+ | * [[American Invitational Mathematics Examination]] | ||
+ | * [[Mathematics competition resources]] | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 15:31, 8 June 2020
Problem
Find the value of if , , is an arithmetic progression with common difference 1, and .
Solution
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 .
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 |