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प्रश्न
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.
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उत्तर
We know that, `"a"_("r" + 1) = "AR"^(("r + 1) - 1)` = ARr;
Where ar = rth 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 ARr, ARr+6 and ARr+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]
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