A) \[a=\frac{7}{4},\,r=\frac{3}{7}\]
B) \[a=\frac{3}{2},\,r=\frac{1}{2}\]
C) \[a=2,\,r=\frac{3}{8}\]
D) \[a=3,\,r=\frac{1}{4}\]
Correct Answer: D
Solution :
Here\[\frac{a}{1-r}=4\,\,\text{and}\,\,ar=\frac{3}{4}\]. Dividing these, \[r(1-r)=\frac{3}{16}\]or \[16{{r}^{2}}-16r+3=0\] or \[(4r-3)(4r-1)=0\] \[r=\frac{1}{4},\frac{3}{4}\,\,\text{and }a=3,\,1\] so \[(a,r)=\left( 3,\frac{1}{4} \right)\,,\,\left( 1,\frac{3}{4} \right)\].
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