Question

(i) Find the 37th term of the AP $6,7\frac{3}{4},9\frac{1}{2},11\frac{1}{4},...$
(ii) Find the 25th term of the AP $5,4\frac{1}{2},4,3\frac{1}{2},3,...$

Answer 1

(i) The given AP is $6,7\frac{3}{4},9\frac{1}{2},11\frac{1}{4},...$
First term, a = 6 and common difference, d $=7\frac{3}{4}-6\Rightarrow \frac{31}{4}-6\Rightarrow \frac{31-24}{4}=\frac{7}{4}$
$$\begin{gathered} \mathrm{Now},T_{37}=a+(37-1)d=a+36d \\ =6+36\times \frac{7}{4}=6+63=69 \\ \therefore {37}^{th}\mathrm{term}=69 \end{gathered}$$

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