A) \[{{P}^{3}}\]
B) \[{{P}^{2}}a\]
C) \[P{{a}^{2}}\]
D) \[{{a}^{3}}\]
Correct Answer: B
Solution :
[b] \[\frac{{{S}_{nx}}}{{{S}_{x}}}=\frac{\frac{nx}{2}[2a+\left( nx-1 \right)d]}{\frac{x}{2}[2a+\left( x-1 \right)d]}\] \[=\frac{n[(2a-d)+nxd]}{\left( 2a-d \right)+xd}\] For \[\frac{{{S}_{nx}}}{{{S}_{x}}}\]to be independent of x \[2a-d=0\] \[\therefore \,\,\,2a=d\] now, \[{{S}_{p}}=\frac{P}{2}[2a+(P-1)d]={{P}^{2}}a\]
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