A) \[0\]
B) \[2\]
C) \[11\]
D) \[29\]
Correct Answer: A
Solution :
Let a and d be the first term and common difference cf the AP, then \[19\times {{T}_{12}}=18\times {{T}_{11}}\] \[19(a+11d)=18(a+10d)\] \[\Rightarrow \] \[19\,a+209\,d=18\,a+180d\] \[\Rightarrow \] \[a+29\,d=0\] \[\Rightarrow \] \[a+(30-1)d=0\] \[\Rightarrow \] \[{{T}_{30}}=0\]You need to login to perform this action.
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