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

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

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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