Question

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

 

  

Answer 1

Let a be the first term and d be the common difference.

We know that, n^th term = a_n = a + (n − 1)d

According to the question,
 
a₇ =  $$\frac{1}{9}$$

⇒ a + (7 − 1)d = $$\frac{1}{9}$$

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

Also, a₉ =  $$\frac{1}{7}$$ 

⇒ a + (9 − 1)d = $$\frac{1}{7}$$

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

On Subtracting (1) from (2), we get
8d − 6d =  $$\frac{1}{7}-\frac{1}{9}$$

⇒ 2d = $$\frac{9-7}{63}$$

⇒ 2d = $$\frac{2}{63}$$

⇒ d = $$\frac{1}{63}$$
⇒ a = $$\frac{1}{9}-\frac{6}{63}$$          [From (1)]
⇒ a =   $$\frac{7-6}{63}$$

⇒ a = $$\frac{1}{63}$$

 

∴ a₆₃ = a + (63 − 1)d
        = $$\frac{1}{63}+\frac{62}{63}$$

 

= $$\frac{63}{63}$$   = 1

Thus, (63)^rd term of the given A.P. is 1.