Question

If the seventh term of an A.P. is $\frac{1}{9}$ and its ninth term is $\frac{1}{7}$, find its $63rd$ term.

Answer 1

Step 1: Compute the value of $d$:

Given that $a_7=\frac{1}{9}$ and $a_9=\frac{1}{7}$

We know that the $nth$ term is ${\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}–1)\mathrm{d}$. That is

$a_7=a+(7-1)d$

$\frac{1}{9}=a+6d$. . . . . . $\left(1\right)$

and

$a_9=a+(9-1)d$

$\frac{1}{7}=a+8d$. . . . . . $\left(2\right)$

Subtract equation $\left(2\right)-\left(1\right)$. That is

$\frac{1}{7}-\frac{1}{9}=(a+8d)-(a+6d)$

$\frac{2}{63}=a+8d-a-6d$

$\frac{2}{63}=2d$

$\mathrm{d}=\frac{2}{63}\times \frac{1}{2}$

$\mathrm{d}=\frac{1}{63}$

Step 2: Compute the value of $a$:

Using equation $\left(1\right)$ we can write that

$\frac{1}{9}=a+6\left(\frac{1}{63}\right)$

$$\begin{gathered} a=\frac{7-6}{63} \\ a=\frac{1}{63} \end{gathered}$$

Step 3: Compute the required term:

$a_{63}=a+62d$ . . . . . . $\left(3\right)$

By putting the value of $a$ and $d$ in the equation $\left(3\right)$ we get

$$\begin{gathered} a_{63}=\frac{1}{63}+62\left(\frac{1}{63}\right) \\ {a}_{63}=\frac{1}{63}+\frac{62}{63} \\ {a}_{63}=\frac{63}{63} \\ {a}_{63}=1 \end{gathered}$$

Hence the $63rd$ term is $1$.