Question

Find the 25th term of the AP $5,4\frac{1}{2},4,3\frac{1}{2},3,...$

Answer 1

The given AP is $5,4\frac{1}{2},4,3\frac{1}{2},3,...$
First term = 5
Common difference$=4\frac{1}{2}-5\Rightarrow \frac{9}{2}-5\Rightarrow \frac{9-10}{2}=-\frac{1}{2}$
$$\begin{gathered} \therefore a=5\mathrm{and}d=-\frac{1}{2} \\ \mathrm{Now},T_{25}=a+(25-1)d=a+24d \\ =5+24\times \left(-\frac{1}{2}\right)=5-12=-7 \\ \therefore {25}^{th}\mathrm{term}=-7 \end{gathered}$$