If a1, a2, a3, ..., ar are in G.P., then prove that the determinant is independent of r.


Given: a1, a2…, ar are in G.P

We know that, ar+1 = AR(r+1)-1 = ARr …(i)


[an = arn-1, where a = first term and r = common ratio]


where A = First term of given G.P


and R = common ratio of G.P


…[from(i)]


Taking ARr, ARr+6 and ARr+10 common from R1, R2 and R3 respectively, we get



If any two columns (or rows) of a determinant are identical (all corresponding elements are same), then the value of determinant is zero.


Here, R1 and R2 are identical.



Hence Proved


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