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Bunuel
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The operation ==> is defined by x==> y = x + (x + 1) + (x + 2) ... +y. 28 Oct 2021, 03:15
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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) ?

A. 2400
B. 2500
C. 2600
D. 2700
E. 2800
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E
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The operation ==> is defined by x==> y = x + (x + 1) + (x + 2) ... +y. 28 Oct 2021, 04:32
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Bunuel
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) ?

A. 2400
B. 2500
C. 2600
D. 2700
E. 2800
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x?y = x + (x + 1) + (x + 2) ... + y MEANS sum of all consecutive integers from x to y, both inclusive.

100?150 = Average *51 = \(\frac{100+150}{2}*51=125*51=6375\)
125?150 = Average *26 = \(\frac{125+150}{2}*26=275*13=3575\)
Difference = 6375-3575 = 2800.


We could also take 100?150-125?150 to be 100+101+…125+126+…..150-(125+126…..+150)=100+101+….+124=100?124
\(\frac{100+124}{2}*25=112*25=2800\)


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The operation ==> is defined by x==> y = x + (x + 1) + (x + 2) ... +y. 30 Nov 2021, 09:30
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Bunuel
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) ?

A. 2400
B. 2500
C. 2600
D. 2700
E. 2800
Show more
------------------------------------
We can also solve it using AP (Arithmetic Progression).
100?150 = 100+101+102+....+125+126+127+.....+148+149+150_________(1)
125?150=125+126+127+.....+148+149+150______________(2)

(1) - (2)= 100+101+102+....+124 (The rest gets cancelled)

n=(124-100)+1 =25
Sum = (n/2) [2a + (n-1) * d]
a = 100
d=1

Substituting the values, we'll get 2800.

Hope this helps! :)
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The operation ==> is defined by x==> y = x + (x + 1) + (x + 2) ... +y. 03 Dec 2024, 19:15
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x?y = x + (x + 1) + (x + 2) ... + y and we need to find the value of (100?150) - (125?150)

x?y = x + (x + 1) + (x + 2) ... + y
=> Sum of all the numbers starting from the number before "?" to the number ending with the number after "?"

(100?150) = Sum of all the numbers from 100 to 150 = 100 + 101 + ... + 149 + 150

(100?150) - (125?150) = ( 100 + 101 + ... + 149 + 150) - (125 + 126 + ... + 149 + 150)
= 100 + 101 + ... + 123 + 124 + [ (125 + 126 +.... + 150) - (125 + 126 +.... + 150) ]
= 100 + 101 + ... + 123 + 124

Now, this is an Arithmetic series with
  • First term = 100
  • Number of terms = 124-100 + 1 = 25
  • Last term = 124
=> Sum = Number of terms * Average of first and last term
[ Watch this video to MASTER Arithmetic Series ]

= 25 * (\(\frac{100 + 124}{2}\))
= 25 * 112
= 2800

So, Answer will be E
Hope it helps!

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