A) \[\frac{3{{a}^{2}}+b}{2}\]
B) \[\frac{3{{a}^{2}}-b}{2}\]
C) \[\frac{b-3{{a}^{2}}}{2}\]
D) \[\frac{{{b}^{2}}-3{{a}^{2}}}{2}\]
Correct Answer: B
Solution :
Let the three terms be \[\alpha ,\beta \] and\[\gamma \]. According to question, \[a=\frac{\alpha +\beta +\gamma }{3}\] ?..(i) and \[b=\frac{{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}}{3}\] ?..(ii) We have to find \[\frac{\alpha \beta +\beta \gamma +\alpha \gamma }{3}\] Squaring (i), we get \[9{{a}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+2[\alpha \beta +\alpha \gamma +\beta \gamma ]\] Using (ii), \[9{{a}^{2}}=3b+2[\alpha \beta +\alpha \gamma +\beta \gamma ]\] \[\Rightarrow \] \[\frac{9{{a}^{2}}-3b}{2}=\alpha \beta +\alpha \gamma +\beta \gamma \] Thus, \[\frac{\alpha \beta +\alpha \gamma +\beta \gamma }{3}=\frac{3{{a}^{2}}-b}{2}\]
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