Given x,y,z are in arithmetic progression, hence 2y=x+z, also ∣3454252kxyz∣=0
Applying R2→R1+R3−2R2
⇒∣30542k−62kx0z∣=0
⇒(k−62)(3z−5x)=0
⇒(k−62)=0 or (3z−5x)=0
If 3z−5x=0
⇒3(x+2d)−5x=0
⇒x=3d (Not possible)
⇒k=62
⇒k2=72.
JEE Main 2021 — Mathematics Algebra
If x,y,z are in arithmetic progression with common difference d,x=3d, and the determinant of the matrix [3454252kxyz] is zero, then the value of k2 is
Held on 17 Mar 2021 · Verified 6 Jul 2026.
72
12
36
6
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