nth term from end of GP

The nth term from end of GP is given by

nth term from the end = l $\left ( \frac{1}{r} \right )^{n - 1}$
where, l = last term
r = common ratio
n = number of terms
Theorem : Prove that the nth term from the end of geometric progression with last term 'l' and the common ratio 'r' is given by
$a_{n} = \left ( \frac{1}{r} \right )^{n - 1}$

Proof : The terms of G.P. from the beginning are a,ar,a$r^{2}$,...,a$r^{n-1}$
If we write this in the reverse order then it will be
a$r^{n-1}$,..., a$r^{2}$,ar,a
First term = a$r^{n-1}$ which is actually a last term
So, let us consider as l = a$r^{n-1}$
Common ratio = $\frac{ar}{ar^{2}} = \frac{1}{r}$

∴ nth term from the end = l $\left ( \frac{1}{r} \right )^{n - 1}$

Examples on finding nth term from end of GP

1) Find 4th term from the end of the G.P. 3,6,12,24,...,3072.
Solution : The given sequence is in G.P with the
last term = l = 3072 and common ratio = r = $\frac{6}{3}$ = 2 and n = 4
nth term from the end = l $\left ( \frac{1}{r} \right )^{n - 1}$

4th term from the end = 3072 $\left ( \frac{1}{2} \right )^{4 - 1}$

= 3072 $\times \frac{1}{8}$

∴ 4th term from the end of G.P = 384

2) Find 6th term from the end of the G.P. 5,10,20,40,...,5120.
Solution : The given sequence is in G.P with the
last term = l = 5120 and common ratio = r = $\frac{10}{5}$ = 2 and n = 6
nth term from the end = l $\left ( \frac{1}{r} \right )^{n - 1}$

6th term from the end = 5120 $\left ( \frac{1}{2} \right )^{6 - 1}$

= 3072 $\times \frac{1}{32}$

∴ 4th term from the end of G.P = 160

3) The last term is $\frac{4}{81}$ and common ratio is $\frac{-1}{3}$. Find the seventh term from the end.
Solution : Here, last term = l = $\frac{4}{81}$ and common ratio = r = $\frac{-1}{3}$ and
n = 7
nth term from the end = l $\left ( \frac{1}{r} \right )^{n - 1}$

7th term from the end = $\frac{4}{81} \left ( \frac{1}{\frac{-1}{3}} \right )^{7 - 1}$

= $\frac{4}{81} \times( -3)^{6}$

= $\frac{4}{81} \times 729$

= 36
∴ 7th term from the end of G.P = 36


11th grade math

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