A) 81
B) 243
C) 343
D) 324
Correct Answer: B
Solution :
[b] Given \[\frac{ar({{r}^{10}}-1)}{r-1}=18\] ...(1) Also \[\frac{\frac{1}{ar}\left( 1-\frac{1}{{{r}^{10}}} \right)}{1-\frac{1}{r}}=6\] Or \[\frac{1}{a{{r}^{11}}}.\frac{({{r}^{10}}-1)r}{r-1}=6\] Or \[\frac{1}{{{a}^{2}}{{r}^{11}}}.\frac{ar({{r}^{10}}-1)}{r-1}=6\] ...(2) From (1) and (2) \[\frac{1}{{{a}^{2}}{{r}^{11}}}.\times 18=6\] Or \[{{a}^{2}}{{r}^{11}}=3\] Now \[P={{a}^{10}}{{r}^{11}}={{({{a}^{2}}{{r}^{11}})}^{5}}={{3}^{5}}=243\]
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