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