A) \[-\frac{3}{2}\]
B) \[\frac{3}{2}\]
C) 2
D) 3
Correct Answer: B
Solution :
Let three terms of G.P. are \[a,\ ar,\ a{{r}^{2}}\]. Then \[a+ar+a{{r}^{2}}=19\Rightarrow a[1+r+{{r}^{2}}]=19\] ?..(i) \[a\ .\ ar\ .\ a{{r}^{2}}=216\Rightarrow {{a}^{3}}{{r}^{3}}=216\Rightarrow ar=6\] ?..(ii) Dividing (ii) by (i), \[\frac{6}{r}+\frac{6}{r}r+\frac{6}{r}{{r}^{2}}=19\Rightarrow \frac{6}{r}+6+6r=19\] \[\Rightarrow {{r}^{2}}-\frac{13}{6}r+1=0\]. Hence\[r=\frac{3}{2}\].
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