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man singh verma

· started a discussion

· 1 Months ago

solution (Question:538412.0)

sir, 10 11 12 hi kyu mana h . give reason
answer ko explain karo

Question:
The sum of the squares of 3 consecutive positive numbers is 365. The sum of the numbers is?
Options:
A) 30
B) 33
C) 36
D) 45
Solution:
102 + 112 + 122 = 100 + 121 + 144 =365 

therefore, Required sum = 10 + 11 + 12 = 33

Knowledge Expert

· commented

· 1 Months ago

Dear student,
⇒x^2+(x+1)^2+(x+2)^2
= 365
⇒x^2+x^2+1+2x+x^2+4+4x= 365
⇒3x^2+6x=360
⇒x^2+2x−120=0
⇒(x−10)(x+12)=0
⇒x=10
Sum of numbers :=10+11+12=33

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