Question

The sum of the first $n$ terms of a sequence is $\frac{7^n-6^n}{6^n}$, Find its $n^{th}$ term. Determine whether the sequence is A.P. or G.P

Answer 1

$S_n=\frac{7^n-6^n}{6^n}$
$=\frac{7^n}{6^n}-\frac{6^n}{6^n}={\left[\frac{7}{6}\right]}^n-1$
$S_{n-1}={\left[\frac{7}{6}\right]}^{n-1}-1$
$t_n=S_n-S_{n-1}$
$={\left[\frac{7}{6}\right]}^n-1-[{\left(\frac{7}{6}\right)}^{n-1}-1]$
$={\left[\frac{7}{6}\right]}^n-{\left[\frac{7}{6}\right]}^{n-1}$
$={\left[\frac{7}{6}\right]}^{n-1}[\frac{7}{6}-1]$
$\therefore t_n=\frac{1}{6}{\left[\frac{7}{6}\right]}^{n-1}$
Now, $\frac{t_{n+1}}{t_n}=\frac{\frac{1}{6}{\left[\frac{7}{6}\right]}^n}{\frac{1}{6}{\left[\frac{7}{6}\right]}^{n-1}}$
$={\left[\frac{7}{6}\right]}^{n-(n-1)}=\frac{7}{6}$
$\therefore \frac{7}{6}$ is constant.
$\therefore$ The given sequence is a G.P.
$\therefore$ The given sequence is a G.P., $t_n=\frac{1}{6}{\left[\frac{7}{6}\right]}^{n-1}$