Correct Answer - Option 3 : 2
Concept:
For a sequence a1, a2, a3 …. an is an G.P.
Common ratio , r = \( \rm a_2 \over a_1\) = \( \rm a_3 \over a_2\) = \( \rm a_n \over a_{n-1}\)
Calculation:
The given general term is,
Tn = \(\rm \frac{5^n}{2^{n -1}}\)
Putting the value of n= 1, 2, 3... we get,
T1 = 5
T2 = \(\rm 5^2 \over 2^{2-1}\) = \(25\over2\)
T3 = \(\rm 5^3 \over 2^{3-1}\) = \(125\over4\)
So the G.P series becomes, 5, \(25\over2\), \(125\over4\)
Where the 1st term is 5 and common ratio is \(25\over2\ \times\ 5\) = \(5\over2\)
So, \(\rm 1st \ term\ \over Common \ ratio\) = \(\rm \frac{5}{\frac{5}{2}}\)= 2