G.P. ain't hard

Let the first term be $a$ , and the common ratio be $r$

Given that:

$T_3=(T_1)^2\\ ar^2 = a^2\\ a^2-ar^2 = 0\\ a(a-r^2)=0\\ a=0 \;\;a=r^2$

Now, we know that $T_2 = ar = 8$ . If $a=0, ar=0 \neq 8$ . Therefore, we reject $a=0$

Substitute $a=r^2$ into $T_2$ :

$ar=8\\ r^2(r)=8\\ r^3 = 8\\ r=2\\ a=r^2=2^2=4$

$T_6 = ar^5 = 4(2^5) = 4(32) = \boxed{128}$

In a Geometric Progression, $a_2 = a_1r$ , $a_3 = a_1r^2$ , $a_4 = a_1r^3$ and so on. We know that $a_2 = \sqrt{a_1a_3}$ . So we have, $a_2 = \sqrt{a_1(a_1)^2} = \sqrt{(a_1)^3}$ . Substituting $8$ to $a_2$ , we get $a_1 = 4$ . Then the common ratio is $\frac{a_2}{a_1} = \frac{8}{4} = 2$ . From here we can solve for the sixth term, $a_6 = a_1r^5 = 4(2^5) = 128$ .

The first term=x the second term=x.r the third=x.r^2 the first term . the third=the second term ^2=64 as the third=first ^2 third . first=x . x^2=64 x^3=64 x=4 so the ratio=2 so the sixth term=4.2^5=128