nth term of GP with first term a and ratio r is:

To find the $n$-th term of a Geometric Progression (GP), we observe the pattern of multiplication starting from the first term $a$. 1. **First term ($T_1$):** By definition, this is $a$. We can write this as $a \cdot r^0$. 2. **Second term ($T_2$):** Multiply the first term by the common ratio $r$: $$T_2 = a \cdot r^1$$ 3. **Third term ($T_3$):** Multiply the second term by $r$: $$T_3 = (a \cdot r) \cdot r = a \cdot r^2$$ 4. **Generalizing the pattern:** - For $n = 1$, the power of $r$ is $0$. - For $n = 2$, the power of $r$ is $1$. - For $n = 3$, the power of $r$ is $2$. Notice that the exponent of $r$ is always one less than the term number $n$. Therefore, for the $n$-th term ($T_n$), the power of $r$ must be $(n-1)$. **Final Formula:** $$T_n = a \cdot r^{n-1}$$