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

Given: a₁, a₂…, a_r are in G.P

We know that, a_r+1 = AR^(r+1)-1 = AR^r …(i)

[∵a_n = ar^n-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 AR^r, AR^r+6 and AR^r+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