Correct Answer - Option 3 : 24
Concept :
Let us consider sequence a1, a2, a3 …. an is a G.P.
- Common ratio , r = \(\rm \frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}}= ... = \frac{a_{n}}{a_{n-1}}\)
-
\(\rm n^{th}\) term of G.P is an = arn-1
- Sum of n terms = s = ; where r >1
- Sum of n terms = s = ; where r <1
- Sum of infinite GP = \(\rm s_{\infty }= \frac{a}{1-r}\) ; |r| < 1
Calculation :
Here 5th term of G.P is 96
i.e a5 = ar5-1
⇒ a5 = ar4
⇒ 96 = ar4 ____( i )
Given: 8th term is 768
⇒ a8 = ar7
768 = ar7 ____(ii)
Divide eqn. (ii) by eqn. (i) , we get
8 = r3
⇒ r = 2 .
Putting this in eqn. (i) , we get
a = 6 .
We know that , \(\rm n^{th}\) term of G.P , an = arn-1
So, a3 = 6× 23-1
⇒ a3 = 24 .
The correct option is 3.