Q7 of 105 Page 139

Find the 31st term of an AP whose 11thterm is 38 and the 16th term is 73

Given

11thterm = 38


⇒ a + (11-1)d = 38 → 1


16th term = 73


⇒ a + (16-1)d = 73 → 2


By subtracting the both equations we will get ‘d’


(a +10d) – (a+15d) = 38 -73


-5d = -35


d = 7


By substituting “d” in equation 1


a +10d = 38


a + 10×7 = 38


a = -32


a31 = a + (31-1) d


= -32 + 30×7


= 178


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