Question

if seventh term of an ap is 1/9 and ninth term is 1/7 find 63 term

Answer 1

Suppose a be the first term and d be the common difference of an A.P.
So seventh term = $t_7=a+\left(7-1\right)d=\frac{1}{9}$
$\Rightarrow a+6d=\frac{1}{9}......\left(\mathrm{i}\right)$
And 9th term = $t_9=a+\left(9-1\right)d=\frac{1}{7}$
$\Rightarrow a+8d=\frac{1}{7}......\left(\mathrm{ii}\right)$
Subtracting (i) from (ii) we get;
$$\begin{gathered} a+8d-a-6d=\frac{1}{7}-\frac{1}{9} \\ \Rightarrow 2d=\frac{9-7}{63} \\ \Rightarrow 2d=\frac{2}{63} \\ \Rightarrow d=\frac{1}{63} \end{gathered}$$
Putting the value of d in either of the above equation , we get;
$$\begin{gathered} a+6\times \frac{1}{63}=\frac{1}{9} \\ \Rightarrow a+\frac{2}{21}=\frac{1}{9} \\ \Rightarrow a=\frac{1}{9}-\frac{2}{21} \\ \Rightarrow a=\frac{7-6}{63}=\frac{1}{63} \end{gathered}$$
So 63rd term = $a+\left(63-1\right)d=\frac{1}{63}+62\times \frac{1}{63}=1$
Therefore 63rd term of an A.P. is 1.