Question

Find the fourth term in each of the following series:
(1) $2,2\frac{1}{2},3\frac{1}{3},....$
(2) $2,2\frac{1}{2},3,.....$
(3) $2,2\frac{1}{2},3\frac{1}{8},....$

Answer 1

$\left(I\right)$ $2,2\frac{1}{2},3\frac{1}{3},...$

$=\frac{2}{1},\frac{5}{2},\frac{10}{3},...$

$⟹$These are in the form of

$T_n=\frac{2^n+(n-1)}{n}$

$\therefore T_4=\frac{2^4+(4-1)}{4}=\frac{16+3}{4}=\frac{19}{4}$

$\left(II\right)$ $2,2\frac{1}{2},3,...$

This is an AP with

$a=2;d=\frac{1}{2}$

$\therefore T_4=a+(4-1)d$

$=2+3\times \frac{1}{2}=2+\frac{3}{2}=\frac{7}{2}=3\frac{1}{2}$

$\left(III\right)$ $2,2\frac{1}{2},3\frac{1}{8},...$

$⟹2,\frac{5}{2},\frac{25}{8},...$

This is GP with,

$a=2$ and $r=\frac{5}{2}\times \frac{1}{2}=\frac{5}{4}$

$\therefore T_4=a{r}^{n-1}$

$=2\times {\left(\frac{5}{4}\right)}^{4-1}=2\times {\left(\frac{5}{4}\right)}^3=\frac{2\times 125}{64}=\frac{125}{32}$