A) 6
B) 5
C) 4
D) 3
E) 2
Correct Answer: D
Solution :
Let a be first term and d be common difference of AP. Then, according to question \[(11th+12th+13th)term=141\] \[\Rightarrow \]\[a+10d+a+11d+a+12d=141\] \[\Rightarrow \]\[3a+33d\,=141\] ?(i) and\[(21th+22th+23th)term=261\] \[\Rightarrow \]\[a+20d+a+21d+a+22d=261\] \[\Rightarrow \] \[3a+63d=261\] ? (ii) Solving Eqs. (i) and (ii), we get \[a=3,\text{ }d=4\] \[\therefore \] First term = 3You need to login to perform this action.
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