Question

If a₁, a₂, a₃, ..., ar are in G.P., then prove that the determinant `|("a"_("r" + 1), "a"_("r" + 5), "a"_("r" + 9)),("a"_("r" + 7), "a"_("r" + 11), "a"_("r" + 15)),("a"_("r" + 11), "a"_("r" + 17), "a"_("r" + 21))|` is independent of r.

Answer 1

We know that, `"a"_("r" + 1) = "AR"^(("r + 1) - 1)` = AR^r;

Where a_r = r^th term of G.P.

A = First term of G.P.

And R = common ratio of G.P.

∴ `|("a"_("r" + 1), "a"_("r" + 5), "a"_("r" + 9)),("a"_("r" + 7), "a"_("r" + 11), "a"_("r" + 15)),("a"_("r" + 11), "a"_("r" + 17), "a"_("r" + 21))| = |("AR"^"r", "AR"^("r" + 4), "AR"^("r" + 8)),("AR"^("r" + 6), "AR"^("r" + 10), "AR"^("r" + 14)),("AR"^("r" + 10), "AR"^("r" + 16), "AR"^("r" + 20))|`

[Taking AR^r, AR^r+6 and AR^r+10 common from R₁, R₂ and R₃, respectively]

= `"AR"^"r" * "AR"^("r" + 6) * "AR"^("r" + 10) |(1, "AR"^4, "AR"^8),(1, "AR"^4, "AR"^8),(1, "AR"^6, "AR"^10)|`

= 0  ....[As R₁ and R₂ are identical]