Difference between revisions of "2020 AMC 10A Problems/Problem 8"
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| − | Looking at the numbers, you see that every set of <math>4</math> has <math>3</math> positive numbers and 1 negative number. Calculating the sum of the first couple sets gives us <math>2+10+18...+394</math>. Clearly, this pattern is an arithmetic sequence. By using the formula we get <math>\frac{2+394}{2}\cdot 50=\boxed{\ | + | Looking at the numbers, you see that every set of <math>4</math> has <math>3</math> positive numbers and 1 negative number. Calculating the sum of the first couple sets gives us <math>2+10+18...+394</math>. Clearly, this pattern is an arithmetic sequence. By using the formula we get <math>\frac{2+394}{2}\cdot 50=\boxed{\textbf{(B) }9900}</math> |
-ViratKohli2018 | -ViratKohli2018 | ||
Revision as of 23:05, 17 March 2020
Contents
Problem
What is the value of
Solution 1
Looking at the numbers, you see that every set of 
has 
positive numbers and 1 negative number. Calculating the sum of the first couple sets gives us 
. Clearly, this pattern is an arithmetic sequence. By using the formula we get 
-ViratKohli2018
Solution 2
Split the even numbers and the odd numbers apart. If we group every 2 even numbers together and add them, we get a total of 
. Summing the odd numbers is equivalent to summing the first 100 odd numbers, which is equal to 
. Adding these two, we obtain the answer of 
.
Solution 3 (bashy)
We can break this entire sum down into integer bits, in which the sum is 
, where 
is the first integer in this bit. We can find that the first sum of every sequence is 
, which we plug in for the 
bits in the entire sequence is 
, so then we can plug it into the first term of every sequence equation we got above 
, and so the sum of every bit is , and we only found the value of , the sum of the sequence is 
.-middletonkids
Solution 4
Another solution involves adding everything and subtracting out what is not needed. The first step involves solving 
. To do this, we can simply multiply 
and 
and divide by 
to get us 
. The next step involves subtracting out the numbers with minus signs. We actually have to do this twice, because we need to take out the numbers we weren’t supposed to add and then subtract them from the problem. Then, we can see that from to , incrementing by , there are numbers that we have to subtract. To do this we can do times 
divided by , and then we can multiply by , because we are counting by fours, not ones. Our answer will be 
, but remember, we have to do this twice. Once we do that, we will get 
. Finally, we just have to do 
, and our answer is .
— 
Solution 5
In this solution, we group every 4 terms. Our groups should be: 
, 
, 
, ... 
. We add them together to get this expression: 
. This can be rewritten as 
. We add this to get 
. ~Baolan
Solution 6
We can split up this long sum into groups of four integers. Finding the first few sums, we have that , , and . Notice that this is an increasing arithmetic sequence, with a common difference of 
. We can find the sum of the arithmetic sequence by finding the average of the first and last terms, and then multiplying by the number of terms in the sequence. The first term is 
, or , the last term is 
, or 
, and there are 
or terms. So, we have that the sum of the sequence is 
, or . ~Arctic_Bunny
Video Solution
~IceMatrix
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
