Answer :

Formula Used.


an = a + (n–1)d


a2 = a + (2–1)d


a2 = a + d


If 2nd term of A.P is given as 1


Then,


a + d = 1


we get a = 1–d ......eq 1


a12 = a + (12–1)d


a12 = a + 11d


If 12th term of A.P is given as –9


Then,


a + 11d = –9


we get a = –9–11d ......eq 2


Equating both eq 1 and eq 2


We get ;


1–d = –9–11d


11d–d = –9–1


10d = –10


d = = –1


Putting d in eq 1 we get ;


a = 1–(–1)


= 2


As a = 2 and d = –1


Then;


a1 = a + (n–1)d = 2 + (1–1)(–1) = 2


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


a3 = a + (n–1)d = 2 + (3–1)(–1) = 2 + (–1)×2 = 2–2 = 0


a4 = a + (n–1)d = 2 + (4–1)(–1) = 2 + (–1)×3 = 2–3 = –1


an = a + (n–1)d = 2 + (n–1)(–1) = 2 + 1–n = 3–n


The A.P is 2, 1, 0, –1……, 3–n


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