Question

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

Answer 1

Given A.P is

$\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4},.....$

first term of this A.P is $a_1=\frac{3}{4}$

second term of this A.P is $a_2=\frac{5}{4}$

common difference of an A.P is given by

$(i.e.,d=a_{n+1}-a_n)$

putting n=1 in above equation

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

hence common difference for this A.P is $d=\frac{2}{4}$

the nth term of an A.P is given by.

$a_n=a_1+(n-1)d$

$⟹a_n=\frac{3}{4}+(n-1)\frac{2}{4}$

$⟹a_n=\frac{3}{4}+\frac{2}{4}n-\frac{2}{4}$

$⟹a_n=\frac{1}{4}+\frac{2}{4}..eq\left(1\right)$

for finding 9th term $\left(a_9\right)$of this A.P put n=9 in above eq(1)

$⟹a_9=\frac{1}{4}+9\times \frac{2}{4}$

$⟹a_9=\frac{1}{4}+\frac{18}{4}$

$⟹a_9=\frac{19}{4}$