Sequences and Series - General term of a G.P.

Find the $10^{th}$ and $n^{th}$ terms of the G.P. $5, 25, 125, \dots$

Given G.P. is $5, 25, 125, \dots$ First term $a = 5$. Common ratio $r = \frac{25}{5} = 5$. Using the formula for the $n^{th}$ term: $a_n = a r^{n-1}$. $a_n = 5 \cdot (5)^{n-1} = 5^1 \cdot 5^{n-1} = 5^{1+n-1} = 5^n$. For the $10^{th}$ term, substitute $n = 10$: $a_{10} = 5^{10}$.

Identify the first term $a$ and common ratio $r$. Substitute these values into the general term formula $a r^{n-1}$ and simplify using laws of exponents.

Which term of the G.P. $2, 8, 32, \dots$ is $131072$?

Here, $a = 2$ and $r = \frac{8}{2} = 4$. Let the $n^{th}$ term be $a_n = 131072$. $a r^{n-1} = 131072$ $2 \cdot (4)^{n-1} = 131072$ $(4)^{n-1} = \frac{131072}{2}$ $(4)^{n-1} = 65536$ Since $65536 = 4^8$, we have: $4^{n-1} = 4^8$ Equating exponents: $n - 1 = 8 \implies n = 9$.

Set the general term formula equal to the given value. Divide by $a$ and express both sides as powers of the same base (the common ratio) to solve for $n$.

In a G.P., the $3^{rd}$ term is $24$ and the $6^{th}$ term is $192$. Find the $10^{th}$ term.

Let $a$ be the first term and $r$ be the common ratio. $a_3 = a r^2 = 24$ --- (1) $a_6 = a r^5 = 192$ --- (2) Dividing equation (2) by (1): $\frac{a r^5}{a r^2} = \frac{192}{24}$ $r^3 = 8 \implies r = 2$ Substitute $r = 2$ into (1): $a(2)^2 = 24 \implies 4a = 24 \implies a = 6$ Now, find the $10^{th}$ term: $a_{10} = a r^9 = 6 \cdot (2)^9 = 6 \cdot 512 = 3072$.

Create a system of two equations using the $n^{th}$ term formula for the given terms. Divide the equations to eliminate $a$ and solve for $r$, then find $a$. Finally, use $a$ and $r$ to calculate the required term.