Question

If five G.M. 's are inserted between $486$ and $\dfrac{2}{3}$then the fourth G.M. will be
A. ${\text{4}}$
B. ${\text{6}}$
C. ${\text{12}}$
D. ${\text{- 6}}$

Answer 1

Hint: The ratio of terms of the geometric progression of the consecutive terms remains the same. Here, we will use the formula for the nth term of the series and with the given values of first and the sixth term will find the fourth term of the series.

Complete step by step answer:
Let us assume the first five GM’s be given as –
${G_1},{G_2},{G_3},{G_4},{G_5}$
So, the given series can be represented as –
$486,{G_1},{G_2},{G_3},{G_4},{G_5},\dfrac{2}{3}$
Above series suggest that the first term be $a = 486$
Seventh term $ = \dfrac{2}{3}$
Place the standard formula in the above expression –
$a{r^6} = \dfrac{2}{3}$
Place the known values in the above expression –
$486\,{r^6} = \dfrac{2}{3}$
Make the required term “r” the subject –
${r^6} = \dfrac{2}{{3 \times 486}}$

Remove common factors from the numerator and the denominator –
${r^6} = \dfrac{1}{{3 \times 243}}$
The term in the denominator can be expressed in the form of power and exponent –
${r^6} = \dfrac{1}{{{3^6}}}$
The above expression can be re-written as –
${r^6} = {\left( {\dfrac{1}{3}} \right)^6}$
Common powers from both the sides of the equation cancels each other.
$r = \dfrac{1}{3}$
Now, fourth GM can be given by ${G_4} = a{r^4}$
Place the known values in the above expression –
${G_4} = 486{\left( {\dfrac{1}{3}} \right)^4}$
Simplify the above expression by removing the common factors from the numerator and the denominator.
$\therefore {G_4} = 6$

Hence, option B is correct.

Note: Power and exponent is a way of expressing the same factors in the short form. Always remember when the bases are the same , powers are added when the terms are in the multiplication whereas when the terms are in division the powers are subtracted. Be good in multiples and get the factors of the terms.