A) \[abc\]
B) \[2abc\]
C) \[3abc\]
D) \[4abc\]
Correct Answer: B
Solution :
Since,\[b,\,\,\,a\]and\[c\]are in\[AP\] \[\therefore \] \[2a=b+c\] ? (i) and\[b,\,\,\,{{G}_{1}},\,\,\,{{G}_{2}}\]and\[c\]are in\[GP\] \[\therefore \] \[{{G}_{1}}=br,\,\,{{G}_{2}}=b{{r}^{2}}\]and\[c=b{{r}^{3}}\] where \[r\] be the common ratio of\[GP\]. Now,\[G_{1}^{3}+G_{2}^{3}={{(br)}^{3}}+{{(b{{r}^{2}})}^{3}}\] \[={{b}^{3}}{{r}^{3}}+{{b}^{3}}{{r}^{6}}\] \[={{b}^{3}}\left( \frac{c}{b} \right)+{{b}^{3}}{{\left( \frac{c}{b} \right)}^{2}}\] \[={{b}^{2}}c+b{{c}^{2}}=bc(b+c)=2abc\] [from Eq. (i)] \[\therefore \] \[G_{1}^{3}+G_{2}^{3}=2abc\]
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