Question

# Let a>0,d>0. Find the value of the determinant ∣∣ ∣ ∣ ∣ ∣ ∣ ∣∣1a1a(a+d)1(a+d)(a+2d)1a(a+d)1(a+d)(a+2d)1(a+2d)(a+3d)1(a+2d)1(a+2d)(a+3d)1(a+3d)(a+4d)∣∣ ∣ ∣ ∣ ∣ ∣ ∣∣

A
Δ=4d4/[a(a+d)2(a+2d)2(a+3d)2(a+4d)]
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B
Δ=4d2/[a(a+d)2(a+2d)2(a+3d)2(a+4d)]
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C
Δ=4d4/[a(a+d)2(a+2d)2(a+3d)2(a+4d)]
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D
None of these.
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Solution

## The correct option is A Δ=4d4/[a(a+d)2(a+2d)2(a+3d)2(a+4d)]Given a>0,d>0Let △=∣∣ ∣ ∣ ∣ ∣∣1a1a(a+d)1(a+d)(a+2d)1a+d1(a+d)(a+2d)1(a+2d)(a+3d)1a+2d1(a+2d)(a+3d)1(a+3d)(a+4d)∣∣ ∣ ∣ ∣ ∣∣Taking 1a(a+d)(a+2d) common from R1,1(a+d)(a+2d)(a+3d) from R2,1(a+2d)(a+3d)(a+4d) from R3⇒△=1a(a+d)2(a+2d)3(a+3d)2(a+4d)∣∣ ∣ ∣∣(a+d)(a+2d)a+2da(a+2d)(a+3d)a+3da+d(a+2d)(a+4d)a+4da+2d∣∣ ∣ ∣∣⇒△=1a(a+d)2(a+2d)3(a+3d)2(a+4d)△1Where △1=∣∣ ∣ ∣∣(a+d)(a+2d)a+2da(a+2d)(a+3d)a+3da+d(a+3d)(a+4d)a+4da+2d∣∣ ∣ ∣∣Applying R2→R2−R1,R3→R3−R2⇒△1=∣∣ ∣ ∣∣(a+d)(a+2d)(a+2d)a(a+2d)(2d)dd(a+3d)(2d)dd∣∣ ∣ ∣∣Applying R3→R3−R2⇒△1=∣∣ ∣ ∣∣(a+d)(a+2d)(a+2d)a(a+d)2ddd2d200∣∣ ∣ ∣∣Expanding along R3, we get△1=∣∣∣a+2dd2d20∣∣∣⇒△1=(2d2)(d)(a+2d−a)=4d2∴△=4d3a(a+d)2(a+2d)3(a+4d)2(a+4d)

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