Difference between revisions of "1984 AIME Problems/Problem 1"
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== Problem == | == Problem == | ||
− | Find the value of <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 == | |
− | |||
− | |||
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} = \boxed{ | + | 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{93}</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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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: | 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{ | + | <math> \frac{137 - 97(49) + 2(49)^2}{2} = \fbox{93} </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{93}</math>. | ||
+ | |||
+ | Or, since the sum of the odd elements is 44, then the sum of the even terms must be <math>\fbox{93}</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 can split <math>a_1+a_2+a_3+\ldots+a_{98}</math> into two parts: | ||
+ | <cmath>a_1+a_2+a_3+\ldots+a_{49}</cmath> and <cmath>a_{50}+a_{51}+a_{52}+\ldots+a_{98}</cmath> | ||
+ | Note that each term in the second expression is <math>49</math> greater than the corresponding term, so, letting the first equation be equal to <math>x</math>, we get <math>a_1+a_2+a_3+\ldots+a_{98}=137=2x+49^2 \implies x=\frac{137-49^2}{2}</math>. Calculating <math>49^2</math> by sheer multiplication is not difficult, but you can also do <math>(50-1)(50-1)=2500-100+1=2401</math>. We want to find the value of <math>x+\frac{49 \cdot 50}{2}=x+49 \cdot 25=x+1225</math>. Since <math>x=\frac{137-2401}{2}</math>, we find <math>x=-1132</math>. <math>-1132+1225=\boxed{93}</math>. | ||
+ | |||
+ | - PhunsukhWangdu | ||
+ | |||
+ | == Solution 5 == | ||
+ | |||
+ | Since we are dealing with an arithmetic sequence, | ||
+ | <cmath>a_2+a_4+a_6+a_8+\ldots+a_{98} = 49a_{50}</cmath> | ||
+ | We can also figure out that | ||
+ | <cmath>a_1+a_2+a_3+\ldots+a_{98} = a_1 + 97a_{50} = 137</cmath> | ||
+ | <cmath>a_1 = a_{50}-49 \Rightarrow 98a_{50}-49 = 137</cmath> | ||
+ | Thus, <math>49a_{50} = \frac{137 + 49}{2} = \boxed{93}</math> | ||
+ | |||
+ | ~kempwood | ||
− | == | + | == Video Solution by OmegaLearn == |
− | + | https://youtu.be/re8DTg-Lbu0?t=234 | |
− | + | ~ pi_is_3.14 | |
== See also == | == See also == | ||
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* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
− | [[Category:Intermediate Algebra Problems]] | + | [[Category: Intermediate Algebra Problems]] |
Latest revision as of 15:33, 24 July 2024
Contents
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 is 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
Solution 5
Since we are dealing with an arithmetic sequence,
We can also figure out that
Thus,
~kempwood
Video Solution by OmegaLearn
https://youtu.be/re8DTg-Lbu0?t=234
~ pi_is_3.14
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 |