Question

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

Answer 1

Hint: In this question it is given that the seventh term of an A.P is $\dfrac{1}{9}$ and its ninth term is $\dfrac{1}{7}$, so we have to find its $63^{rd}$ term. So to find the solution we need the expression for $n^{th}$ term, i.e, $t_{n}=a+\left( n-1\right) d$.......(1)
Where a = first term of A.P and d = common difference.
So by the above equation we will find a and d, and after that we can easily find the $63^{rd}$ term i.e, $t_{63}$.

Complete step-by-step solution:
The seventh term of this Arithmetic progression is $\dfrac{1}{9}$.
Therefore, by (1) we can write,
$t_{7}=\dfrac{1}{9}$
$\Rightarrow a+\left( 7-1\right) d=\dfrac{1}{9}$
$\Rightarrow a+6d=\dfrac{1}{9}$.............(2)
Also the ninth term is $\dfrac{1}{7}$.
i.e, $t_{9}=\dfrac{1}{7}$
$a+\left( 9-1\right) d=\dfrac{1}{7}$
$a+8d=\dfrac{1}{7}$..................(3)

Now by subtracting (2) from (3), we get,
$\left( a+8d\right) -\left( a+6d\right) =\dfrac{1}{7} -\dfrac{1}{9}$
$\Rightarrow a+8d-a-6d=\dfrac{9-7}{63}$
$\Rightarrow 2d=\dfrac{2}{63}$
$\Rightarrow d=\dfrac{1}{63}$
Now by putting the value of d in equation (2), we get,
$a+6\times \dfrac{1}{63} =\dfrac{1}{9}$
$\Rightarrow a+\dfrac{6}{63} =\dfrac{1}{9}$
$\Rightarrow a=\dfrac{1}{9} -\dfrac{6}{63}$
$\Rightarrow a=\dfrac{7-6}{63}$
$\Rightarrow a=\dfrac{1}{63}$

Therefore, the $63^{rd}$ term is,
$t_{63}=a+\left( 63-1\right) d$
$=\dfrac{1}{63} +62\times \dfrac{1}{63}$
$=\dfrac{1}{63} +\dfrac{62}{63}$=1
Therefore, the $63^{rd}$ term is 1.

Note: Always remember that if any series is in the form of A.P then the difference of consecutive two terms is constant and this difference is called common difference(d). Also in some books you will find that the $n^{th}$ term is written as $a_{n}$ instead of $t_{n}$, but both of them define $n^{th}$ term.