Answer :

Formula Used.


an = a + (n–1)d


a7 = a + (7–1)d


a7 = a + 6d


If 7th term of A.P is given as 12


Then,


a + 6d = 12


we get a = 12–6d ......eq 1


a12 = a + (12–1)d


a12 = a + 11d


If 12th term of A.P is given as 72


Then,


a + 11d = 72


we get a = 72–11d ......eq 2


Equating both eq 1 and eq 2


We get ;


12–6d = 72–11d


11d–6d = 72–12


5d = 60


d = = 12


Putting d in eq 1 we get ;


a = 12–6×12


= 12–72 = –60


As a = –60 and d = 12


Then;


a1 = a + (n–1)d = –60 + (1–1)(12) = –60


a2 = a + (n–1)d = –60 + (2–1)(12) = –60 + (12)×1 = –48


a3 = a + (n–1)d = –60 + (3–1)(12) = –60 + (12)×2 = –36


a4 = a + (n–1)d = –60 + (4–1)(12) = –60 + (12)×3 = –24


an = a + (n–1)d = –60 + (n–1)(12) = 12n – 72


The A.P is –60, –48, –36, –24……, 12n – 72


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