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Four different integers form an increasing A.P. If one  of these numbers is equal to the sum of the squares of  the other three numbers, then the numbers are

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a
−2,−1,0,1
b
0,1,2,3
c
−1,0,1,2
d
None of these

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detailed solution

Correct option is C

Let the numbers be a−d,a,a+d,a+2dwhere,a,d∈Z and d>0Given: (a−d)2+a2+(a+d)2=a+2d⇒ 2d2−2d+3a2−a=0∴ d=121±1+2a−6a2Since d is positive integer,∴1+2a−6a2>0⇒1−760∴ d=1. Hence, the numbers are −1,0,1,2.


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