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AMC 10 Math Olympiad USA Math Olympiad

Sum of digits | AMC-10A, 2020 | Problem 8

Try this beautiful problem from Algebra, based on Sum of digits from AMC-10A, 2020. You may use sequential hints to solve the problem

Try this beautiful problem from Algebra based on sum of digits

Sum of digits – AMC-10A, 2020- Problem 8


What is the value nof

\(1+2+3-4+5+6+7-8+……+197+198+199-200\)?

  • \(9800\)
  • \(9900\)
  • \(10000\)
  • \(10100\)
  • \(10200\)

Key Concepts


Algebra

Arithmetic Progression

Series

Check the Answer


Answer: \(9900\)

AMC-10A (2020) Problem 8

Pre College Mathematics

Try with Hints


The given sequence is \(1+2+3-4+5+6+7-8+……+197+198+199-200\). if we look very carefully then notice that \(1+2+3-4\)=\(2\), \(5+6+7-8=10\),\(9+10+11-12=18\)…..so on.so \(2,10,18……\) which is in A.P with common difference \(8\). can you find out the total sum which is given….

can you finish the problem……..

we take four numbers in a group i.e \((1+2+3-4)\),\((5+6+7-8)\),\((9+10+11-12)\)……,\((197+198+199-200)\). so there are \(\frac{200}{4}=50\) groups. Therefore first term is\((a)\)= \(2\) ,common difference\((d)\)=\(8\) and numbers(n)=\(50\). the sum formula of AP is \(\frac{n}{2}\{2a+(n-1)d\}\)

can you finish the problem……..

\(\frac{n}{2}\{2a+(n-1)d\}\)=\(\frac{50}{2}\{2.8+(50-1)8\}\)=\(9900\)

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