`T_(n) = (5n -2)` (given)
`rArr T_(1) = (5 xx 1 -2) = 3 "and"T_(2) = (5 xx 2-2) = 8.`
Thus, we have
(i) first term = 3.
(ii) common difference `= (T_(2) -T_(1)) = (8-3) = 5.`
(iii) 19th term = a +(19-1) d, where a = 3 and d = 5.
` = (3 + 18 xx 5 ) = 93.`