The sum of the first 5 terms of a geometric progression is the same as the sum of the first 7 terms of the same progression. If the sum of the first 9 terms is 24, then the 4th term of the progression is

Sum of first $$n$$ terms of a G.P. is given by the formula, $$S$$=$$\dfrac{a\cdot\left(r^n-1\right)}{r-1}$$ where $$a$$ is the first term and $$r$$ is the common ratio.

So according to question,

$$\dfrac{a.\left(r^5-1\right)}{r-1}=a\cdot\dfrac{\left(r^7-1\right)}{r-1}$$

or, $$r^5=r^7$$

or, $$r^5(1-r^2)=0$$

or, $$r=0,1,-1$$

But $$r$$ can't be 1 as then sum of terms will be not defined (in denominator $$r-1$$ is there so for $$r=1$$, sum will be not defined)

Also, $$r$$ can't be 0 as then all the other terms except the first term will be zero.

So, $$r=-1$$

Now, sum of first 9 terms=24

or, $$\dfrac{a.\left(r^9-1\right)}{r-1}=\dfrac{a\left(-1-1\right)}{\left(-1-1\right)}=24$$

or, $$a=24$$

So, fourth term = $$a\cdot r^{4-1}=a\cdot r^3=24\left(-1\right)^3=-24$$