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.