If $G$ is the GM of the product of $r$ sets of observations with geometric means $G_{1}, G_{2}, . . ., G_{r}$ respectively, then $G$ is equal to

log G_{1} + log G_{2} + . . . + log G_{r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

G_{1} . G_{2} . . . . . G_{r}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

log G_{1} . log G_{2} . . . . . log G_{r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $G_{1} . G_{2} . . . . . G_{r}$
Let x be the product of numbers $x_{1}$ , $x_{2}$ , ..., $x_{r}$
corresponding to $r$ sets of observations i.e. $x = (x_{1} . x_{2} . . . . x_{r})$
we have $log x = log x_{1} + log x_{2} + . . . + log x_{r}$
$∴$ $\sum log x = \sum log x_{1} + \sum log x_{2} + . . . + \sum log x_{r}$
$∴$ $\frac{\sum log x}{n} = \frac{\sum log x_{1}}{n} + \frac{\sum log x_{2}}{n} + . . . + \frac{\sum log x_{r}}{n}$
$log G = log (G_{1}) + log (G_{2}) + . . . + log (G_{r})$
$log G = log (G_{1} G_{2} . . . G_{r})$
$∴$ $G = G_{1} . G_{2} . G_{3} . . . G_{r}$

Suggest Corrections

Similar questions

G_{1}, G_{2}

G

G

G_{1}, G_{2}, . . . . . . . G_{n}

If n geometric means between a and b be $G_{1}, G_{2}, . . . . G_{n}$ and a geometric mean of a and b be G, then the true relation is

G_{1}

G_{2}

G_{3}

G_{k}

G_{1}, G_{2}

\frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}

Join BYJU'S Learning Program