Correct Answer  Option 3 : 24
Concept :
Let us consider sequence a_{1}, a_{2}, a_{3} …. a_{n} is a G.P.
 Common ratio , r = \(\rm \frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}}= ... = \frac{a_{n}}{a_{n1}}\)

\(\rm n^{th}\) term of G.P is a_{n} = ar^{n1}
 Sum of n terms = s = ; where r >1
 Sum of n terms = s = ; where r <1
 Sum of infinite GP = \(\rm s_{\infty }= \frac{a}{1r}\) ; r < 1
Calculation :
Here 5^{th }term of G.P is 96
i.e a_{5} = ar^{51}
⇒ a_{5} = ar^{4}
⇒ 96 = ar^{4 }( i )
Given: 8th term is 768
⇒ a8 = ar7
768 = ar^{7} (ii)
Divide eqn. (ii) by eqn. (i) , we get
8 = r^{3 }
⇒ r = 2 .
Putting this in eqn. (i) , we get
a = 6 .
We know that , \(\rm n^{th}\) term of G.P , a_{n} = ar^{n1}
So, a_{3} = 6× 2^{3}^{1}
⇒ a_{3} = 24 .
The correct option is 3.