A) \[b=\frac{a+c}{2}\]
B) \[{{a}^{2}}=bc\]
C) \[{{b}^{2}}=ac\]
D) \[{{c}^{2}}=ab\]
Correct Answer: C
Solution :
Let first term of G.P. \[=A\] and common ratio \[=r\] We know that \[{{n}^{th}}\]term of G.P. = \[A{{r}^{n-1}}\] Now \[{{t}_{4}}=a=A{{r}^{3}},\ {{t}_{7}}=b=A{{r}^{6}}\]and \[{{t}_{10}}=c=A{{r}^{9}}\] Relation \[{{b}^{2}}=ac\]is true because \[{{b}^{2}}={{(A{{r}^{6}})}^{2}}={{A}^{2}}{{r}^{12}}\] and \[ac=(A{{r}^{3}})(A{{r}^{9}})={{A}^{2}}{{r}^{12}}\] Aliter: As we know, if \[xy+2{{y}^{2}}+yz=xy+xz+{{y}^{2}}+yz\] in A.P., then \[=2{{n}^{2}}+5n-2{{n}^{2}}+4n-2-5n+5=4n+3\] terms of a G.P. are always in G.P., therefore, \[a,\ b,\ c\] will be in G.P. \[i.e.\]\[2,\ 5,\ 8,\ 11,\ 14=40\].
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