If the determinant ∣∣
∣∣abat−bbcbt−c210∣∣
∣∣=0, if a,b,c are in
A
A.P.
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B
G.P.
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C
H.P.
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D
k=1/2
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Solution
The correct option is BG.P. ∣∣
∣∣abat−bbcbt−c210∣∣
∣∣=0 Expanding it along third row, ⇒2[b(bt−c)−c(at−b)]−1[a(bt−c)−b(at−b)]=0 ⇒2t(b2−ac)−(b2−ac)=0 ⇒(2t−1)(b2−ac)=0 ⇒t=12 or b2=ac If b2=ac then a,b,c∈ G.P.