Difference between revisions of "1984 AIME Problems/Problem 1"
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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: | ||
<cmath>a_1+a_2+a_3+\ldots+a_{49}</cmath> and <cmath>a_{50}+a_{51}+a_{52}+\ldots+a_{98}</cmath> | <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>. | + | 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 | - PhunsukhWangdu | ||
Revision as of 15:28, 10 August 2021
Contents
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 .
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 |