Question

Find the ${103}^{th}$ term of the $A.P.4,4\frac{1}{2},5,5\frac{1}{2},6,..$

Answer 1

${103}^{th}terma=4,d=\frac{1}{2}$

We know that

$T_n=a+\left(n-1\right)d$

Therefore,

$T_{103}=4+\left(103-1\right)\times \frac{1}{2}=4+\frac{102}{2}=4+51=55$

Hence, this is the answer.