\(\frac {A_4}{r^3}. \frac {A_4}{r} . A_4r . A_4r^3 = \frac {1}{1296}\)
\(A_4 = \frac 16\)
\(A_2 = \frac {7}{36} - \frac 16\)
\(A_2= \frac {1}{36}\)
So, \(A_6 + A_8 + A_{10} = 1 + 6 + 36\)
\(= 43\)
So, the correct option is (C): \(43\)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A geometric progression is the sequence, in which each term is varied by another by a common ratio. The next term of the sequence is produced when we multiply a constant to the previous term. It is represented by: a, ar1, ar2, ar3, ar4, and so on.
Important properties of GP are as follows:
If a1, a2, a3,… is a GP of positive terms then log a1, log a2, log a3,… is an AP (arithmetic progression) and vice versa