Question

If x, y, z are in A.P., a, b, c are in H.P and ax, by, cz are in G.P then​

Answer 1

EXPLANATION.

=> x, y, z are in Ap.

=> y = x + z / 2 .......(1)

=> a, b, c are in H. P.

=> b = 2ac / a + c .......(2)

=> ax, by, cz are in G. P

=> by = √ax X cz ........(3)

From equation (1) and (2) and (3) we get,

put the value of y and b in equation (3)

$= > \: \: \bold{\frac{2ac}{a \: + c} \times \frac{x + z}{2} = \sqrt{ax \: \times cz} }$

$= > \: \: \bold{\frac{(x + z) { {}^{2} } }{(a + c) {}^{2} } \times (ac) {}^{2} = ac \: \times xz}$

$= > \: \: \bold{\frac{(x \: + z) {}^{2} }{xz} = \frac{(a \: + c) {}^{2} }{ac} }$

$= > \: \bold{\frac{ {x}^{2} }{xz} + \frac{ {z}^{2} }{xz} + \frac{2xz}{xz} = \frac{ {a}^{2} }{ac} + \frac{ {c}^{2} }{ac} + \frac{2ac}{ac}}$

$= > \: \: \frac{x}{z} + \frac{z}{x} + 2 = \frac{a}{c} + \frac{c}{a} + 2$

$= > \: \: \bold{ \frac{x}{z} + \frac{z}{x} = \frac{a}{c} + \frac{c}{a} }$

Answer 2

x, y, z are in A.P.

$\implies \: \: \bold{y = x + \frac{z}{2} ............(i) }$

a, b, c are in H. P.

$\implies \: \: \bold{b = \frac{2ac}{a + c} ............(ii)}$

ax, by, cz are in G.P.

$\implies \: \: \bold{by = \sqrt{ax} \times cz .........(iii)}$

From equation (i) and (ii) and (iii),

Put the value of y and b in equation (iii),

$\implies{\frac{2ac}{a \: + c} \times \frac{x + z}{2} = \sqrt{ax \: \times cz} }$

$\implies{\frac{(x + z) { {}^{2} } }{(a + c) {}^{2} } \times (ac) {}^{2} = ac \: \times xz}$

$\implies{\frac{(x \: + z) {}^{2} }{xz} = \frac{(a \: + c) {}^{2} }{ac} }$

$\implies \: {\frac{ {x}^{2} }{xz} + \frac{ {z}^{2} }{xz} + \frac{2xz}{xz} = \frac{ {a}^{2} }{ac} + \frac{ {c}^{2} }{ac} + \frac{2ac}{ac}}$

$\implies\frac{x}{z} + \frac{z}{x} + 2 = \frac{a}{c} + \frac{c}{a} + 2$

$\implies{ \frac{x}{z} + \frac{z}{x} = \frac{a}{c} + \frac{c}{a} }$