Question

If the 9th term of an ap is 1/7 and the 7th term os 1/9 find the 63rd term

Answer 1

`a_(63)` = 1

Explanation:

The n th term of an arithmetic progression is

`a_(n)` = a + (n - 1)d

where a is the first term and d the common difference

use the 9 th term and 7 th term to find a and d

`a_(9)` = a + 8d =
`(1)/(7)` → (1)

`a_(7)` = a + 6d =
`(1)/(9)` → (2)

Subtract (2) from (1) term by term

2d =
`(2)/(63)` ⇒ d =
`(1)/(63)`

Substitute this value into (2) and solve for a

a +
`(6)/(63)` =
`(1)/(9)`

a =
`(1)/(9)` -
`(6)/(63)` =
`(1)/(63)`

Hence

`a_(63)` =
`(1)/(63)` +
`(62)/(63)` = 1