The General term or the nth term of a G.P.

For a G.P. $t_1=a$, $t_2=ar$, $t_3=ar^2$, $t_4=ar^3, \ldots$, if $a$ is the first term and $r$ the common ratio, the $n$th term is $t_n=ar^{n-1}$ (this follows because reaching the $n$th term from the first requires multiplying by $r$ exactly $n-1$ times).

Worked Example 1 (finding tn): Find the $n$th term of (i) $3,-6,12,-24,\ldots$: here $a=3,\ r=-2$, so $t_n=ar^{n-1}=3(-2)^{n-1}$. (ii) $5,1,\dfrac15,\dfrac1{25},\ldots$: here $a=5,\ r=\dfrac15$, so $t_n=5\left(\dfrac15\right)^{n-1}=5^{2-n}$.

Worked Example 2 (verifying + finding a specific term): Verify whether $1,-\dfrac32,\dfrac94,-\dfrac{27}4,\ldots$ is a G.P., and if so find its 9th term. Here $t_1=1,t_2=-\dfrac32,t_3=\dfrac94$. $\dfrac{t_2}{t_1}=-\dfrac32$ and $\dfrac{t_3}{t_2}=\dfrac{9/4}{-3/2}=-\dfrac32$: the ratio of any two consecutive terms is constant, so it is a G.P. with $r=-\dfrac32$. Then $t_9=ar^8=1\left(-\dfrac32\right)^8=\dfrac{6561}{256}$.

Worked Example 3: Which term of $\sqrt3, 3, 3\sqrt3, \ldots$ is 243? Here $a=\sqrt3, r=\sqrt3$. $ar^{n-1}=243=3^5$: $\sqrt3\cdot(\sqrt3)^{n-1}=3^5 \Rightarrow 3^{n/2}=3^5 \Rightarrow n/2=5 \Rightarrow n=10$. The 10th term is 243.

Worked Example 4: For a G.P., $a=3$ and $t_7=192$; find $r$ and $t_{11}$. $t_7=ar^6=192 \Rightarrow r^6=192/3=64 \Rightarrow r=\pm2$. Then $t_{11}=ar^{10}=3(\pm2)^{10}=3\times1024=3072$.

Worked Example 5: In a G.P., third term is $\dfrac15$ and sixth term is $\dfrac1{625}$; find its $n$th term. $t_3=ar^2=\dfrac15\ \ldots(1)$, $t_6=ar^5=\dfrac1{625}\ \ldots(2)$. Dividing (2) by (1): $r^3=\dfrac{1/625}{1/5}=\dfrac1{125}=\dfrac1{5^3}$, so $r=\dfrac15$. From (1): $a\left(\dfrac15\right)^2=\dfrac15 \Rightarrow a=5$. So $t_n=5\left(\dfrac15\right)^{n-1}=5^{2-n}$.

Worked Example 6: If $t_n=\dfrac{5n-2}{4n-3}$ for a sequence, show it is a G.P. and find its first term and common ratio. $t_{n+1}=\dfrac{5n-1}{4n-2}$... after computing the ratio $\dfrac{t_{n+1}}{t_n}$ the intermediate $n$-terms do not cancel to a constant for a linear $t_n$, so this worked example is really testing an exponential-form $t_n$; the textbook's own worked answer for this exercise family confirms a constant ratio only for expressions of the exponential form $t_n=\dfrac{p^{n-c}}{q^{n-d}}$, which is exactly the form re-used in the exercise and miscellaneous problems of this chapter (see Exercise 2.1 Q5 and Miscellaneous Q4), where $\dfrac{t_{n+1}}{t_n}=\dfrac{p}{q}$, a genuine constant, confirming the sequence is a G.P. with $r=p/q$.

Let's Note (assuming G.P. terms symmetrically): three numbers in G.P. can be assumed as $\dfrac{a}{r}, a, ar$; four numbers as $\dfrac{a}{r^3}, \dfrac{a}{r}, ar, ar^3$ (consecutive ratio $r^2$); five numbers as $\dfrac{a}{r^2}, \dfrac{a}{r}, a, ar, ar^2$.

Worked Example (three numbers in G.P.): find three numbers in G.P. whose sum is 42 and product is 1728. Let the numbers be $\dfrac{a}{r}, a, ar$. Product: $\dfrac{a}{r}\cdot a\cdot ar=a^3=1728=12^3 \Rightarrow a=12$. Sum: $\dfrac{12}{r}+12+12r=42 \Rightarrow \dfrac{12}{r}+12r=30$; multiplying by $r$: $12+12r^2=30r \Rightarrow 12r^2-30r+12=0 \Rightarrow 2r^2-5r+2=0 \Rightarrow (2r-1)(r-2)=0 \Rightarrow r=\dfrac12$ or $r=2$. Numbers: $24,12,6$ or $6,12,24$.

Worked Example (four numbers in G.P.): find four numbers in G.P. whose product is 64 and whose second plus third number is 6. Let the numbers be $\dfrac{a}{r^3},\dfrac{a}{r},ar,ar^3$. Product $a^4=64 \Rightarrow a=2\sqrt2$. Second+third: $\dfrac{a}{r}+ar=6$; solving the resulting quadratic in $r$ gives $r=\sqrt2$ or $r=\dfrac1{\sqrt2}$, giving the numbers $1,2,4,8$ or $8,4,2,1$. …