If the $4^{t h}, 10^{t h}$ and $16^{t h}$ terms of a GP are x, y and z respectively, then x,y,z are in :

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AGP

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Solution

The correct option is B GP
Let a be the first term and r be the common ratio of the G.P
According to the given condition,
$a_{4} = a r^{3} = x . . . . . (1)$
$a_{10} = a r^{9} = y . . . . . (2)$
$a_{16} = a r^{15} = z . . . . . (3)$
Dividing (2) by (1), we get
$\frac{y}{x} = \frac{a r^{9}}{a r^{3}}$
$\frac{y}{x} = a^{6}$
Dividing (3) by (2), we get
$\frac{z}{y} = \frac{a r^{15}}{a r^{9}}$
$\frac{z}{y} = a^{6}$
$∴ \frac{y}{x} = \frac{z}{y}$
Thus x,y,z are in G.P

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