How Cheenta works to ensure student success?
Explore the Back-Story

# Value of Sum | PRMO - 2018 | Question 16

Try this beautiful Problem based on Value of Sum from PRMO 2018, Question 16.

## Value of Sum - PRMO 2018, Question 16

What is the value of $\sum_{1 \leq i<j \leq 10 \atop i+j=\text { odd }}(i-j)-\sum_{1 \leq i<j \leq 10 \atop i+j=\text { even }}(i-j) ?$

• $50$
• $53$
• $55$
• $59$
• $65$

### Key Concepts

Odd-Even

Sum

integer

Answer:$55$

PRMO-2018, Problem 16

Pre College Mathematics

## Try with Hints

We have to find out the sum . Now substitite $i=1,2,3...9$ and observe the all odd-even cases......

Can you now finish the problem ..........

$i=1 \Rightarrow$$1+(2+4+6+8+10-3-5-7-9) =1-4+10=7 i=2 \Rightarrow$$0 \times 2+(3+5+7+9-4-6-8-10)$
$=-4$

$i=3 \Rightarrow$$1 \times 3+(4+6+8+10-5-7-9) =3-3+10=10 i=4 \Rightarrow$$ 0 \times 4+(5+7+9-6-8-10)=-3$
$i=5 \Rightarrow $$1 \times 5+(6+8+10-7-9)=5-2+10 =13 i=6 \Rightarrow$$ 0 \times 6+(7+9-8-10)=-2$
$i=7 \Rightarrow $$1 \times 7+(8+10-9)=7-1+10=16 i=8 \Rightarrow$$ 0 \times 8+(9-10)=-1$
$i=9 \Rightarrow$$1 \times 9+(10)=19 Can you finish the problem........ Therefore S =(7+10+13+16+19)-(4-3-2-1) =55 ## Subscribe to Cheenta at Youtube Try this beautiful Problem based on Value of Sum from PRMO 2018, Question 16. ## Value of Sum - PRMO 2018, Question 16 What is the value of \sum_{1 \leq i<j \leq 10 \atop i+j=\text { odd }}(i-j)-\sum_{1 \leq i<j \leq 10 \atop i+j=\text { even }}(i-j) ? • 50 • 53 • 55 • 59 • 65 ### Key Concepts Odd-Even Sum integer ## Check the Answer Answer:55 PRMO-2018, Problem 16 Pre College Mathematics ## Try with Hints We have to find out the sum . Now substitite i=1,2,3...9 and observe the all odd-even cases...... Can you now finish the problem .......... i=1 \Rightarrow$$ 1+(2+4+6+8+10-3-5-7-9)$
$=1-4+10=7$
$i=2 \Rightarrow $$0 \times 2+(3+5+7+9-4-6-8-10) =-4 i=3 \Rightarrow$$ 1 \times 3+(4+6+8+10-5-7-9)$
$=3-3+10=10$
$i=4 \Rightarrow$$0 \times 4+(5+7+9-6-8-10)=-3 i=5 \Rightarrow$$1 \times 5+(6+8+10-7-9)=5-2+10$
$=13$
$i=6 \Rightarrow$$0 \times 6+(7+9-8-10)=-2 i=7 \Rightarrow$$1 \times 7+(8+10-9)=7-1+10=16$
$i=8 \Rightarrow$$0 \times 8+(9-10)=-1 i=9 \Rightarrow$$ 1 \times 9+(10)=19$

Can you finish the problem........

Therefore $S =(7+10+13+16+19)$-$(4-3-2-1)$ =$55$

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.