Let a be the first term and d be the common difference of given A.P. Second term, a+d=12 . . . (i) Sum of first nine terms, S9=
9
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(2a+8d)=9(a+4d) Given that S9 is more than 200 and less than 220 ⇒200<S9<220 ⇒200<9(a+4d)<220 ⇒200<9(a+d+3d)<220 Putting value of (a+d) from equation (i) 200<9(12+3d)<220 ⇒200<108+27d<220 ⇒200−108<108+27d−108<220−108 ⇒92<27d<112 Possible value of d is 4 27×4=108 Thus, 92<108<112 Putting value of d in equation (i) a+d=12 a=12−4=8 4th term =a+3d=8+3×4=20