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If $^{2 n} C_{4},^{2 n} C_{5}$ and $^{2 n} C_{6}$ are in A.P, then n is equal to

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Correct option is C

2 2nC5=2nC4+2nC6⇒2= 2nC4 2nC5+ 2nC62n=52n−4+2n−56⇒12(2n−4)=30+(2n−4)(2n−5)⇒2n2−21n+49=0⇒n=7

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If 2nC4,2nC5 and 2nC6 are in A.P, then n is equal to

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If $^{2 n} C_{4},^{2 n} C_{5}$ and $^{2 n} C_{6}$ are in A.P, then n is equal to