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……..

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

## Other useful links

- https://www.cheenta.com/probability-in-coordinates-amc-10a-2003-problem-12/
- https://www.youtube.com/watch?v=XOrePzJWFiE