If (p-1),(p-2),(3p-1)are in Ap,then p is equal to : 1.(4) 2.(-4)3.(2)4.(-2)
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If you write p + 3 in place of p -2 then the answer is 4.p−1,p+3,3p−1 are in A.P.∴2(p+3)=p−1+3p−1⇒2p+6=4p−2⇒2p=8∴p=4
(p -2) – (p -1) = (3p -1) – (p -2)p -2 – p + 1 = 3p – 1 – p + 2-1 = 2p + 12p = -1 -1\xa02p = -2p = -2/2 = -1
But answer is 4
(p -2) – (p -1) = (3p -1) – (p -2)p -2 – p + 1 = 3p – 1 – p + 2-1 = 2p + 12p = -1 -1\xa02p = -2p = -2/2 = -1