A) \[8\]
B) \[12\]
C) \[18\]
D) \[24\]
Correct Answer: C
Solution :
[c] Given a, b, c are in G.P. i.e., \[a,\text{ }ar,\text{ }a{{r}^{2}}\]are in G.P. Now, \[a-2,\text{ }2b\]and \[12c\] are in G.P. or \[a-2,\text{ }2ar,\text{ }12a{{r}^{2}}\]are in G.P. with common ratio 5 (given). Hence, \[5=\frac{2ar}{a-2}=\frac{12a{{r}^{2}}}{2ar}\] \[\Rightarrow \,\,\,\,r=\frac{5}{6}\] and \[a=3\] Hence, sum \[=a+b+c+...\infty =\frac{a}{1-r}=\frac{3}{1-\frac{5}{6}}=18\]You need to login to perform this action.
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