# nth term of a GP

Go back to  'Arithmetic-Geometric'

Consider a GP with first term equal to a and common ratio equal to r. The second term will be $$ar$$, the third term will be $$a{r^2}$$, the fourth term will be $$a{r^3}$$, the tenth term will be $$a{r^9}$$, and so on. Clearly, the nth term will be the product of a and r to the power:

${T_n} = a{r^{n - 1}}$

For example, consider the following GP:

$\frac{1}{4},\,\frac{1}{8},\,\frac{1}{{16}},\,\frac{1}{{32}},\,...$

The first term of this GP is $$a = \frac{1}{4}$$, and its common ratio is $$r = \frac{1}{2}$$. Let us calculate some particular terms of this GP:

\begin{align}&{T_7} = a{r^6} = \left( {\frac{1}{4}} \right) \times {\left( {\frac{1}{2}} \right)^6} = \frac{1}{{{2^8}}}\\&{T_{11}} = a{r^{10}} = \left( {\frac{1}{4}} \right) \times {\left( {\frac{1}{2}} \right)^{10}} = \frac{1}{{{2^{12}}}}\\&{T_{100}} = a{r^{99}} = \left( {\frac{1}{4}} \right) \times {\left( {\frac{1}{2}} \right)^{99}} = \frac{1}{{{2^{101}}}}\end{align}

Remark: Suppose that three terms $$x,y,z$$ are in GP. This means that

$\frac{y}{x} = \frac{z}{y}\,\,\, \Rightarrow \,\,\,{y^2} = xz$

Thus, the square of the middle term is equal to the product of the first and the third term.

Example 1: Consider the following GP:

$2,\,8,\,32,\,...$

Which term of this GP is 131072?

Solution: We observe that for this GP:

$a = 2,\,\,\,r = 4$

Now, let 131072 be the nth term of this GP. We have:

$\begin{array}{l}{T_n} = a{r^{n - 1}} = 131072\\ \Rightarrow \,\,\,2 \times {\left( 4 \right)^{n - 1}} = 131072\\ \Rightarrow \,\,\,n = 9\end{array}$

Example 2: For a particular GP with a positive common ratio, the first term is 1, while the sum of the third and the fifth terms is 90. Find the 10th term of this GP.

Solution: We have $$a = 1$$. Let the common ratio be r. Now, we have:

$\begin{array}{l}{T_3} + {T_5} = 90\\ \Rightarrow \,\,\,a{r^2} + a{r^4} = 90\\ \Rightarrow \,\,\,{r^2} + {r^4} = 90\end{array}$

By observation, it is easy to see that $$r = 3$$. Thus, the 10th term of this GP is

${T_{10}} = a{r^9} = {3^9}$

Example 3: Find all such GPs in which the sum of the first two terms is $$- 4$$, and the fifth term is four times the third term.

Solution: Let a and r be the first term and the common ratio (respectively) of the GP. We have:

$\begin{array}{l}{T_1} + {T_2} = - 4\,\,\, \Rightarrow \,\,\,a + ar = - 4\\{T_5} = 4{T_3}\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,a{r^4} = 4a{r^2}\end{array}$

From the second equation, we obtain $$r = \pm 2$$. If $$r = 2$$, then $$a = - \frac{4}{3}$$. On the other hand, if $$r = - 2$$, then $$a = 4$$. Thus, two different GPs are possible which satisfy the specified constraints:

\begin{align}&r = 2,\,\,\,a = - \frac{4}{3} :\,\,\, - \frac{4}{3},\, - \frac{8}{3},\, - \frac{{16}}{3},...\\&r = - 2,\,\,\,a = 4\,\,\,\,:\,\,\,4,\, - 8,\,16,\, - 32,\,...\end{align}

Example 4: The pth, qth and rth terms of a GP are a, b and c respectively. Prove that:

${a^{q - r}} \times {b^{r - p}} \times {c^{p - q}} = 1$

Solution: If x and y are the first term and the common ratio of the GP, we have:

\begin{align}&a = x{y^{p - 1}},\,\,\,b = x{y^{q - 1}},\,\,\,c = x{y^{r - 1}}\\ &\Rightarrow \,\,\,\left\{ \begin{array}{l}{a^{q - r}} = {\left( {x{y^{p - 1}}} \right)^{q - r}} = {x^{q - r}}{y^{\left( {p - 1} \right)\left( {q - r} \right)}}\\{b^{r - p}} = {\left( {x{y^{q - 1}}} \right)^{r - p}} = {x^{r - p}}{y^{\left( {q - 1} \right)\left( {r - p} \right)}}\\{c^{p - q}} = {\left( {x{y^{r - 1}}} \right)^{p - q}} = {x^{p - q}}{y^{\left( {r - 1} \right)\left( {p - q} \right)}}\end{array} \right.\end{align}

If we multiply these three terms, it is easy to see that the exponents of x will add to 0, and so will the exponents of y. Thus, the product will equal 1.

Example 5: In an AP, the pth, qth, rth and sth terms are in GP. Show that the following terms are in GP:

$\left( {p - q} \right),\,\,\left( {q - r} \right),\,\,\left( {r - s} \right)$

Solution: Suppose that the first term and the common difference of the AP are a and d respectively. We have:

$\begin{array}{l}{T_p} = a + \left( {p - 1} \right)d\\{T_q} = a + \left( {q - 1} \right)d\\{T_r} = a + \left( {r - 1} \right)d\\{T_s} = a + \left( {s - 1} \right)d\end{array}$

It is given that these four terms are in GP. Let R be the common ratio of this GP. Then:

$\begin{array}{l}{T_q} = R{T_p},\,\,\,{T_r} = {R^2}{T_p},\,\,\,{T_s} = {R^3}{T_p}\\ \Rightarrow \,\,\,\left\{ \begin{array}{l}{T_p} - {T_q} = \left( {1 - R} \right){T_p}\\{T_q} - {T_r} = R\left( {1 - R} \right){T_p}\\{T_r} - {T_s} = {R^2}\left( {1 - R} \right){T_p}\end{array} \right.\end{array}$

Clearly, these three terms are in GP, which means that:

$\begin{array}{l}{\left( {{T_q} - {T_r}} \right)^2} = \left( {{T_p} - {T_q}} \right)\left( {{T_r} - {T_s}} \right)\\ \Rightarrow \,\,\,{\left\{ {\left( {q - r} \right)d} \right\}^2} = \left\{ {\left( {p - q} \right)d} \right\}\left\{ {\left( {r - s} \right)d} \right\}\\ \Rightarrow \,\,\,{\left( {q - r} \right)^2} = \left( {p - q} \right)\left( {r - s} \right)\end{array}$

Thus, the terms $$\left( {p - q} \right),\,\,\left( {q - r} \right),\,\,\left( {r - s} \right)$$ are in GP.

Example 6: Four numbers form an increasing GP such that the difference between the first and the third terms is 9, while the difference between the second and the fourth terms is 18. Find these numbers.

Solution: Let the first number be a, and let the common ratio of the GP be r. Thus, the four numbers are:

$a,\,ar,\,a{r^2},\,a{r^3}$

Now, we have:

$\begin{array}{l}a{r^2} - a = 9\,\,\, & \Rightarrow \,\,\,a\left( {{r^2} - 1} \right) = 9\\a{r^3} - ar = 18\,\,\, & \Rightarrow \,\,\,ar\left( {{r^2} - 1} \right) = 18\end{array}$

Dividing these two relations gives $$r = 2$$. Substituting this value of r in any of the two relations gives $$a = 3$$. Thus, the four numbers are:

$3,\,6,\,12,\,24$

Example 7: The first and the nth terms of a GP are a and b respectively. If P is the product of the first n terms, find the value of $${\left( {ab} \right)^n}$$ in terms of P.

Solution: If the common ratio of the GP is r, we have $$b = a{r^{n - 1}}$$. Now, the product of the first n terms is:

\begin{align}&P = a \times ar \times a{r^2} \times a{r^3} \times ... \times a{r^{n - 1}}\\\,\,\,\,\, &\;\;= {a^n} \times {r^{\left( {1 + 2 + 3 + ... + \left( {n - 1} \right)} \right)}}\\\,\,\,\,\, &\;= {a^n} \times {r^{\frac{{n\left( {n - 1} \right)}}{2}}}\end{align}

We now find $${\left( {ab} \right)^n}$$:

\begin{align}&{\left( {ab} \right)^n} = {\left( {a \times a{r^{n - 1}}} \right)^n}\\\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;= {a^{2n}}{r^{n\left( {n - 1} \right)}}\, = {\left( {{a^n}{r^{\frac{{n\left( {n - 1} \right)}}{2}}}} \right)^2} = {P^2}\end{align}

Example 8: Three numbers in GP have a sum of 56. If we subtract 1, 7, 21 from these numbers (respectively), the resulting numbers are in AP. Find the three numbers.

Solution: Let the three numbers be $$a,\,ar,\,a{r^2}$$. Since their sum is 56, we have:

$\begin{array}{l}a + ar + a{r^2} = 56\\ \Rightarrow \,\,\,a\left( {1 + r + {r^2}} \right) = 56 & & ...(i)\end{array}$

According to the problem, the following three numbers are in AP:

$a - 1,\,\,\,ar - 7,\,\,\,a{r^2} - 21$

Thus,

$\begin{array}{l}2\left( {ar - 7} \right) = \left( {a - 1} \right) + \left( {a{r^2} - 21} \right)\\ \Rightarrow \,\,\,2ar - 14 = a\left( {1 + {r^2}} \right) - 22\\ \Rightarrow \,\,\,a\left( {1 - 2r + {r^2}} \right) = 8 & & ...(ii)\end{array}$

Dividing (i) by (ii), we have:

\begin{align}&\frac{{1 + r + {r^2}}}{{1 - 2r + {r^2}}} = 7\\ &\Rightarrow \,\,\,1 + r + {r^2} = 7 - 14r + 7{r^2}\\ &\Rightarrow \,\,\,6{r^2} - 15r + 6 = 0\\ &\Rightarrow \,\,\,2{r^2} - 5r + 2\end{align}

If we factorize this quadratic, we have:

$\begin{array}{l}\left( {2r - 1} \right)\left( {r - 2} \right) = 0\\ \Rightarrow \,\,\,r = \frac{1}{2},\,\,\,r = 2\end{array}$

For the first value of r, a comes out to 32; for the second value, a is 8. Thus, the three numbers are: {8, 16, 32} or {32, 16, 8}.

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