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Find the value of x if 2x + 1, x2 + x +1, 3x2 - 3x +3 are consecutive terms of an AP.
Ans: a2 -a1 = a3 -a2
⇒ x2 + x + 1-2x - 1 = 3x2 - 3x + 3- x2-x-1
x2 - x = 2x2 - 4x + 2
⇒ x2 - 3x + 2 = 0
⇒ (x -1) (x - 2) = 0
⇒ x = 1 or x = 2
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In figure, the incircle of triangle ABC touches the sides BC, CA, and AB at D, E, and F respectively. Show that AF+BD+CE=AE+BF+CD= 1/2 (perimeter of triangle ABC), Ans:
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