Let a be the first term and d be the common difference of the given AP. Then,
`T_(7) = (1)/(9) rArr a +6d = (1) /(9) " "…(i)`
`T_(9) = (1)/(7) rArr a + 8d = (1)/(7) " " ... (ii)`
On subtracting (i) form (ii), we get
`2d = ((1)/(7) - (1)/(9)) = (2)/(63) rArr d = ((1)/(2) xx (2)/(63)) = (1)/(63).`
Putting ` d = (1)/(63)` in (i), we get
`a +(6 xx (1)/(63)) = (1)/(9) rArr a + (2)/(21) = (1)/(9) rArr a = ((1)/(9) - (2)/(21)) = ((7-6)/(63)) = (1)/(63)`
`"Thus,"a = (1)/(63) "and "d = (1)/(63).`
`therefore T_(63) = a + (63 -1)d =(a+62d)`
`= ((1)/(63) + 62 xx (1)/(63)) = ((1)/(63) + (62)/(63)) = 1.`
Hence, 63 rd term of the given AP is 1.
