A) 8
B) 16
C) 20
D) 24
Correct Answer: C
Solution :
Let a be the first term and d be the common difference of given A.P. Second term, a+ d = 12 ...(1) Sum of first nine terms, \[{{S}_{9}}=\frac{9}{2}(2a+8d)=9(a=4d)\] Given that \[{{S}_{9}}\]is more than 200 and less than \[\Rightarrow \]\[200<{{S}_{9}}<220\] \[\Rightarrow \]\[200<9(a+4d)<220\] \[\Rightarrow \]\[200<9(a+d+3d)<220\] Putting value of (a + d) from equation (1) \[200<9(12+3d)<220\] \[\Rightarrow \]\[200<108+27d<220\] \[\Rightarrow \]\[200-108<108+27d-108<220-108\] \[\Rightarrow \]\[92<27d<112\] Possible value of d is 4 \[27\times 4=108\] Thus, 92 < 108 < 112 Putting value of d in equation (1) a + d = 12 a = 12 ? 4 = 8 4th term \[=a+3d=8+3\times 4=20\]
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