Question

Find the 8th term of the sequence :

$\frac{3}{4},1\frac{1}{2},3,..............$

Answer 1

$AP:\frac{3}{4},1\frac{1}{2},3,.............orAP:\frac{3}{4},\frac{3}{2},3,..............orAP:\frac{3}{4},\frac{6}{4},3,.............a=\frac{3}{4}d=a_2-a_1=\frac{6}{4}-\frac{3}{4}=\frac{6-3}{4}=\frac{3}{4}$

$n^{th}term,a_n=a+(n-1)d=a_8=\frac{3}{4}+(8-1)\frac{3}{4}=\frac{3}{4}+7\times \frac{3}{4}=\frac{3}{4}+\frac{21}{4}=\frac{21+3}{4}=\frac{24}{4}=6$