A) \[\frac{1}{2}\]
B) 4
C) 2
D) \[\frac{7}{13}\]
Correct Answer: B
Solution :
\[a=A+6d\] |
\[b=A+10\] |
\[c=A+12\] |
a, b, c are in G.P. |
\[\Rightarrow \] \[{{(A+10d)}^{2}}=(A+6d)(a+12d)\] |
\[\Rightarrow \] \[\frac{A}{d}=-\,14\] |
\[\frac{a}{c}=\frac{A+6d}{A+12d}\] |
\[=\,\frac{6+\frac{A}{d}}{12+\frac{A}{d}}=\frac{6-14}{12-14}=4\] |
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