A) 4, 8, 12, 16
B) 3, 6, 9, 12
C) 4, 7, 10, 13
D) 3, 7, 11, 15
Correct Answer: D
Solution :
Let 1st term of A.P. be a and common difference be d. Now, three middle terms of this A.P. are \[{{a}_{10}},{{a}_{11}}\]and \[{{a}_{12}}\]. According to the question, \[{{a}_{10}}+{{a}_{11}}+{{a}_{12}}=129\] \[\Rightarrow \] \[(a+9d)+(a+10d)+(a+11d)=129\] \[\Rightarrow \] \[3a+30d=129\Rightarrow a+10d=43\] \[\Rightarrow \] \[a=43-10d\] ... (i) Also, last three terms are \[{{a}_{19}},{{a}_{20}}\] and \[{{a}_{21}}\]. \[\therefore \] \[{{a}_{19}}+{{a}_{20}}+{{a}_{21}}=237\] \[\Rightarrow \] \[(a+18d)+(a+19d)+(a+20d)=237\] \[\Rightarrow \] \[3a+57d=237\,\,\Rightarrow \,\,a+19d=79\] \[\Rightarrow \] \[43-10d+19d=79\] (Using (i)) \[\Rightarrow \] \[9d=36\,\,\,\Rightarrow \,\,\,d=4\] and \[a=3\] \[\therefore \] A.P. is 3, 7, 11, 15....
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