Question

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

A
4da(a+d)(a+2d)(a+3d)(a+4d)
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B
4da(a+d)2(a+2d)3(a+3d)2(a+4d)
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C
4d4a(a+d)2(a+2d)3(a+3d)2(a+4d)
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D
\N
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Solution

## The correct option is C 4d4a(a+d)2(a+2d)3(a+3d)2(a+4d)Given, a > 0, d > 0 and let Δ=∣∣ ∣ ∣ ∣ ∣∣1a1a(a+d)1(a+d)(a+2d)1(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)∣∣ ∣ ∣ ∣ ∣∣ 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 ⇒ Δ=1a(a+d)2(a+2d)3(a+3d)2(a+4d) where, Δ′=∣∣ ∣ ∣∣(a+d)(a+2d)(a+2d)a(a+2d)(a+3d)(a+3d)(a+d)(a+3d)(a+4d)(a+4d)(a+2d)∣∣ ∣ ∣∣ Applying R2→R2−R1,R3→R3−R2 ⇒ Δ′=∣∣ ∣ ∣∣(a+d)(a+2d)(a+2d)a(a+2d)(2d)dd(a+3d)(2d)dd∣∣ ∣ ∣∣ Applying R3→R3−R2 Δ′=∣∣ ∣ ∣∣(a+d)(a+2d)(a+2d)a(a+2d)2ddd2d200∣∣ ∣ ∣∣ Expanding along R3, we get Δ′=2d2∣∣∣a+2dadd∣∣∣Δ′=(2d2)(d)(a+2d−a)=4d4∴ Δ=4d4a(a+d)2(a+2d)3(a+3d)2(a+4d)

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