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Q.

If a1, a2, a3, …, a12 are in A.P. and Δ1=a1a5a1a2a2a6a2a3a3a7a3a4 Δ2=a2a10a2a3a3a11a3   a4a3a12a4a5 then Δ1:Δ2=____________ .

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Detailed Solution

Δ1=a12+4a1da1da22+4a2da2da32+4a3da3d,     C3→C3−C2where d is the common difference of A.P.   =da12a11a22a21a32a31+4da1a1da2a2da3a3d=da1−a2a2−a3a3−a1=−2d4Similarly, Δ2=−2d4
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