consecutive integers

[Johnrohit]

the operation --> is defined by x -->y= x+(x+1)+(x+2)...+y. For example, 3-->7=3+4+5+6+7. What is the value of (100 -->150) - (125 -->150)?


[spoiler]OA=2800[/spoiler]

[Matt@VeritasPrep]

Let's get each term clear first, then look for a way to simplify.

100 → 150 = (100 + 101 + 102 + ... + 149 + 150)
123 → 150 = (123 + 124 + 125 + ... + 149 + 150)

So when I subtract the sum in green from the sum in red, everything will cancel except for (100 + 101 + 102 + ... + 121 + 122).

This means that the question is really asking me for the sum of the integers from 100 to 122, inclusive. The formula for this is (Smallest Integer + Greatest Integer)/2 times (Number of Integers), which here is

(100+122)
--------- * 23
2

Or 111 * 23, or 2553. (The OA is wrong, or there was a typo in the question: that answer would be right if the question was (100 → 150) - (125 → 150).)

[Johnrohit]

Sooo sorry matt! thank u for the help! tht was a typo!! edited it! wil check d posts b4 posting it 4m nw on! thanks again

Let's get each term clear first, then look for a way to simplify.

100 → 150 = (100 + 101 + 102 + ... + 149 + 150)
123 → 150 = (123 + 124 + 125 + ... + 149 + 150)

So when I subtract the sum in green from the sum in red, everything will cancel except for (100 + 101 + 102 + ... + 121 + 122).

This means that the question is really asking me for the sum of the integers from 100 to 122, inclusive. The formula for this is (Smallest Integer + Greatest Integer)/2 times (Number of Integers), which here is

(100+122)
--------- * 23
2

Or 111 * 23, or 2553. (The OA is wrong, or there was a typo in the question.)

the operation --> is defined by x -->y= x+(x+1)+(x+2)...+y. For example, 3-->7=3+4+5+6+7. What is the value of (100 -->150) - (123 -->150)?


[spoiler]OA=2800[/spoiler]

[GMATGuruNY]

the operation --> is defined by x -->y= x+(x+1)+(x+2)...+y. For example, 3-->7=3+4+5+6+7. What is the value of (100 -->150) - (125 -->150)?


[spoiler]OA=2800[/spoiler]

100-->150 = 100+101+...+123+124+125+126+...+149+150
125-->150 = .....................................+125+126+...+149+150
When we subtract, all of the values in red cancel (as Matt noted above).
Thus:
Difference = 100+101+....+123+124.

We need to calculate the sum of the CONSECUTIVE INTEGERS from 100 to 124, inclusive.

For any set of consecutive integers:
Number of integers = biggest - smallest + 1
Average = median = (biggest + smallest)/2
Sum = (number)(average).

Here:
Number of integers = 124-100+1 = 25.
Average = (124+100)/2 = 112.
Sum = 25*112 = 2800.

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