A) 55th
B) 60th
C) 65th
D) 70th
Correct Answer: C
Solution :
(c): Here, first term a = 3, common difference \[d=15-3=12\] \[\therefore {{a}_{54}}=a+53d=3+53\times 12\] \[[\because {{a}_{n}}=a+(n-1)d]\] \[=3+636=639\] Let \[{{a}_{k}}\] be 132 more than its 54th term. \[\therefore {{a}_{k}}={{a}_{54}}+132\];\[{{a}_{k}}=639+132\] \[\Rightarrow {{a}_{k}}=771\] \[\Rightarrow a+(k-1)d=771\] \[[\because {{a}_{k}}=a+(n-1)d]\] \[\Rightarrow 3+(k-1)12=771\] \[\Rightarrow 12\,(k-1)=771-3\] \[\Rightarrow k-1=\frac{768}{12}=64\] \[\Rightarrow k=65\] Hence, 65th term is 132 more than its 54th term of an AP.
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