If x is a real number such that x (x^2+1), -x^2/2 , 6 are three consecutive terms of an AP then find the next two consecutive terms of the AP.

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Please find below the solution to the asked query :

x x2+1 , -x22 , 6 are in AP .So ,-x22-x x2+1=6+x22-x2-2x3-2x2=12+x22-x2-2x3-2x=12+x22x3+2x2+2x+12=0x3+x2+x+6=0Factors of 6 are ±1 , ±2 , ±3 , ±6 .For x=-2x3+x2+x+6=-8+4-2+6=0So x+2 is a factor of x3+x2+x+6Now ,x3+x2+x+6=0x3+2x2-x2-2x+3x+6=0x2 x+2 - x x+2+ 3x+2=0x+2 x2-x+3=0x2-x+3=0 have no real roots .So , x=-2 .a=-24+1=-10a+d=-42=-2d=-2+10=8a+3d=-10+38=14a+4d=-10+48=22Next two terms are 14 and 22 .
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