Answer :

Formula Used.


an = a + (n–1)d


a5 = a + (5–1)d


a5 = a + 4d


If 5th term of A.P is given as 17


Then,


a + 4d = 17


we get a = 17–4d ......eq 1


a9 = a + (9–1)d


a9 = a + 8d


If 9th term of A.P is given as 35 + 2nd term


Then,


a2 = a + (1)d


= a + d


a + 8d = 35 + [a + d]


we get


a–a + 8d–d = 35


7d = 35


d = = 5 ......eq 2


Putting d in eq 1 we get ;


a = 17–(4×5)


= 17–20 = –3


As a = –3 and d = 5


Then;


a1 = a + (n–1)d = –3 + (1–1)(5) = –3


a2 = a + (n–1)d = –3 + (2–1)(5) = –3 + (5)×1 = –3 + 5 = 2


a3 = a + (n–1)d = –3 + (3–1)(5) = –3 + (5)×2 = –3 + 10 = 7


a4 = a + (n–1)d = –3 + (4–1)(5) = –3 + (5)×3 = –3 + 15 = 12


an = a + (n–1)d = –3 + (n–1)(5) = –3 + 5n–5 = 5n–8


The A.P is –3, 2, 7, 12, ……, 5n–8


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