for the circuit shown , if i want to calculate gain of the second stage then do i need to consider the output resistance of first stage as the source resistance for the second stage ?
If yes then the output resistance of first stage will be
R_O1 = r_e ll R_E ll r_o = .02513 KΩ
then gain equation will become A_V2 = g_m2(R_C)(r_e2llR_S) = 76.92
but in the solutions they have ignored the effect of R_S and used  A_V2 = g_m2(R_C)
please clarify this

Hi aman gupta,
I am not quite happy with raj`s answer because he speaks about a "battery" (?) of voltage Vo2 (why Vo2?). Therefore, I like to clarify.
The answer is straightforward:
You have a two-stage amplifier (common-collector and common base).
*Calculate the gain Vo1/Vin of the 1st stage considering the pretty low input resistance of the 2nd stage in parallel to Re.
* Calculate the gain of the 2nd stage Vo2/Vo1 without any internal source resistance.
* Multiply both gains resulting in Vo2/Vin
* With other words: Consider Vo1 as an IDEAL voltage source. Otherwise you would consider Re twice in your calculation.
* Explanation: This approach is allowed because of the SUBSTITUTION THEOREM, which unfortunately is not very well known (but, nevertheless, very often applied without knowing it).
This theorem is explained in "Desoer, Kuh: Basic Circuit Theory".

sorry
in the first post i did some mistake.  In that i wrote the formula for voltage gain of common base as Av = (gm/RS)(RcllRL)(re ll RE ll RS)
and for my circuit the second stage is common base for which RE is ∞ and no RL . So gain expression will become Av = (gm2/RS)(Rc)(re ll RS)
i forgot to divide by Rs. But still answer in the textbook for gain of second stage is 153.8 and the factor of 2 is coming because they ignored Rs.
This is a question from electronic circuits analysis & analysis by donald neamen

no it is not solved...........
i try to explain the question once again
what i want to say is in case of cascading if we are calculating the gain of a stage then we have to consider the effects of previous and the next stage.
Now i want to calculate the gain of the 2nd stage which is a common base configuration. So the voltage gain formula for common base amplifier with a non ideal voltage source at the input and a load resistance R_L at the output i know is
Av = (gm/R_S)(RcllR_L)(r_e ll R_E ll R_S)
where g_m is I_CQ/V_T
R_S is the resistance of the voltage source
R_E is the resistance connected between the emitter and ground (small signal model)
r_e = V_T/I_EQ

But when R_S is 0 then gain expression is A_V=g_m(R_C ll R_L )

So in the given circuit the R_O of the first stage should become the R_S for the second stage

Now in the question they have asked to calculate the gain of both stages and the overall gain.
In the solutions they have calculated the gain of the 1st stage considering the input resistance of the 2nd stage as the load of 1st stage. This is fine.
R_L for 2nd stage is infinite

Now to calculate the gain of 2nd stage they are taking R_S = 0 and hence the gain formula for 2nd stage (common base) reduces to A_V2 = g_m2R_C.    This is my doubt that why he can consider R_S for 2nd stage to be 0

got it . thanks  :)