The first and eight terms of a GP. are $x^{-4}$ and $x^{52}$ respectively. If the second term is $x^t$, then t is equal to:

Let a be the first term and r be the common ratio so, general term of GP is $T_n=a{r}^{n-1}$ As given,

${\mathrm{T}}_1={\mathrm{x}}^{-4}=\mathrm{a}$ and ${\mathrm{T}}_8={\mathrm{ar}}^7={\mathrm{x}}^{52}\therefore {\mathrm{ar}}^7={\mathrm{x}}^{52}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}^{-4}{r}^7=x^{52}\Rightarrow r^7=x^{56}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{r}^7={\left(x^8\right)}^7\Rightarrow r=x^8\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{T}_2=a{r}^1=x^{-4}\cdot x^8\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{T}_2=x^4\end{array}$

But ${\mathrm{T}}_2=x^t$

$\Rightarrow {\mathrm{x}}^{\mathrm{t}}={\mathrm{x}}^4\Rightarrow \mathrm{t}=4$