Omitted Progression-2

Let the common ratio of the geometric progression be $r$ . Then we have:

$\begin{aligned} 1296r^2 & =36\\ \implies r^2 & =\frac {36}{1296}=\frac 1{36} \\ r & =\frac 16 \end{aligned}$

The sum of the three missing terms is given below.

$\begin{aligned} S & = a_1+a_3 +a_5 \\ & = \frac {1296} r + 1296r+36r \\ & = 1296\times 6+ \frac {1296} 6 +\frac {36}6 \\ & =7776+216+6 \\ & =\boxed {7998} \end{aligned}$

$\text{Since they are in GP, the common ratio will be the same} \\ a_n = ar^{n-1}; ~ a_1 = ?, ~ a_2 = 1296, ~ a_3 = ?, ~ a_4 = 36, ~ a_5 = ? \\ \dfrac{a_2}{a_1} = \dfrac{a_4}{a_3} \\ \dfrac{1296}{a} = \dfrac{36}{ar^2} \\ 36 = \dfrac{1}{r^2} \\ r = \dfrac{1}{6} \\ a_2 = 1296 \\ ar = 1296 \\ a(\dfrac{1}{6}) = 1296 \\ a = 7776 \\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \text{Therefore, sum of missing terms } S_m \text{, is sum of first terms - sum of known terms}\\ S_m = S_5 - a_2 - a_4 \\ S_m = 7776 \cdot \dfrac{1 - (\dfrac{1}{6})^5}{1-\dfrac{1}{6}} - 1296 - 36 \\ S_m = 7776 \cdot \dfrac{1555}{1296} - 1332 \\ S_m = 9330 - 1332 \\ \boxed{\therefore S_m = 7998}$

$\text{As the given sequence is in Geometric Progression,there should be some 'r'(common ratio), such that:}$ $1296r^2=36$ $\text{So by calculation,}$ $r=\pm\dfrac16$

$\text{But,it is very much clear that if we consider the common ratio as}$ $-\dfrac16,$ $\text{then our final answer,i.e.,the sum of all the missing terms will be negative.But the options are positive.So we take the other case:}$ $\text{As}$ $r=\dfrac16,$ $\text{the last term will be 36r,the middle one 1296r, and the first one will be}$ $\dfrac{1296}r$ . $\text{Adding,we get}$ $\textbf{7998}$ .