Find:9th term of the A.P. $\frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, ………$

Given:

Given A.P. is $\frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, ………$

To do:
We have to find the 9th term of the given A.P.

Solution:

Here,

$a_1=\frac{3}{4}, a_2=\frac{5}{4}, a_3=\frac{7}{4}, a_4=\frac{9}{4}$

Common difference $d=a_2-a_1=\frac{5}{4}-\frac{3}{4}=\frac{5-3}{4}=\frac{2}{4}$

We know that,

nth term $a_n=a+(n-1)d$

Therefore,

9th term $a_{9}=\frac{3}{4}+(9-1)\frac{2}{4}$

$=\frac{3}{4}+8\times\frac{2}{4}$

$=\frac{3}{4}+\frac{16}{4}$

$=\frac{3+16}{4}$

$=\frac{19}{4}$

The 9th term of the given A.P. is $\frac{19}{4}$.