Question

LVI)
(i) The 4th, 7th and 10th terms of a GP. are a, b, c
respectively. Show that b2 = ac.
(ii) If the 4th, 10th and 16th terms of al GP. are x, y, z
respectively, prove that x, y, z are in GP.​

Answer 1

Answer:

Please refer to the attachment...

Answer 2

Answer:

Given -

The 4th, 10th and 16th terms of GP are x, y, z

To Prove -

x, y, z are in GP

Solution -

${ar}^{2} = x$

${ar}^{6} = y$

${a}^{9} = z$

Geometric means of GP is

${b}^{2} = a$

${ar}^{6} = \sqrt{ {ar}^{3} . {ar}^{9} }$

$\sqrt{ {a}^{2} {r}^{12} } = \sqrt{( {ar}^{6} } )^{2}$

$= {ar}^{6}$

${y}^{2} = xz$

Hence, x, y, z are in G. P

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