Prove that $a_{1} a_{3} - a_{1} a_{4}$ is positive, zero or negative according as a $a_{1}, a_{2}, a_{3}, a_{4}$ are in A.P.,G.P or H.P.

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Solution

(i) $a_{1}, a_{2}, a_{3}, a_{4}$ are in G.P
$∴ a_{2} = a_{1} + d, a_{3} = a_{1} + 2 d, a_{4} = a_{1} + 3 d$
$∴ a_{2} a_{3} - a_{1} a_{4} = (a_{1} + d) (a_{1} + 2 d) - a_{1} (a_{1} + 3 d)$
$= (a_{1}^{2} + 3 a_{1} d + 2 d^{2}) - (a_{1}^{2} + 3 a_{1} d) = 2 d^{2} = + i v e$
(ii) $a_{1}, a_{2}, a_{3}, a_{4}$ are in G.P
$∴ \frac{a_{2}}{a_{1}} = \frac{a_{4}}{a_{3}} ∴ a_{2} a_{3} - a_{1} a_{4} = 0$
(iii) $a_{1}, a_{2}, a_{3}, a_{4}$ are in H.P.
$∴ \frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, \frac{1}{a_{4}}$ are in A.P. of common difference D, say
$\frac{1}{a_{4}} - \frac{1}{a_{1}} = 3 D o r a_{1} - a_{4} = 3 D a_{1} a_{4}$ ...(1)
$\frac{1}{a_{2}} = \frac{1}{a_{1}} + D = \frac{1 + a_{1} D}{a_{1}} ∴ a_{2} = \frac{a_{1}}{1 + a_{1} D}$
$\frac{1}{a_{3}} = \frac{1}{a_{4}} - D = \frac{1 - a_{4} D}{a_{4}} ∴ a_{3} = \frac{a_{4}}{1 - a_{4} D}$
$∴ a_{2} a_{3} - a_{1} a_{4} = [\frac{1}{(1 + a_{1} D) (1 - a_{4} D)} - 1]$
$= \frac{a_{1} a_{4} [1 - 1 + (a_{1} - a_{4}) D - a_{1} a_{4} D^{2}]}{1 + (a_{1} - a_{4}) D - a_{1} a_{4} d^{2}}$
Now use (1)
$= \frac{a_{1} a_{4} [- 3 D^{2} a_{1} a_{4} + a_{1} a_{4} D^{2}]}{1 + 3 D^{2} a_{1} a_{4} - a_{1} a_{4} d^{2}}$
$= \frac{(a_{1} a_{4} D^{2})}{1 + 2 D^{2} a_{1} a_{4}} = - i v e$

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