A) \[\frac{1}{2}{{p}^{3}}\]
B) \[mnp\]
C) \[{{p}^{3}}\]
D) \[(m+n){{p}^{2}}\]
Correct Answer: C
Solution :
We have given that, \[{{S}_{n}}={{n}^{2}}p\] and \[{{S}_{m}}={{m}^{2}}p\] Thus, \[{{n}^{2}}p=\frac{n}{2}[2a+(n-1)d]\] \[\Rightarrow \] \[2np=2a+(n-1)d\] ....(i) Similarly, \[2np=2a+(n-1)d\] ....(ii) Subtracting (ii) from (i), we get \[2p(n-m)=(n-m)d\Rightarrow d=2p\] Now, \[2np=2a+(n-1)\times 2p\] [From (i)] \[\Rightarrow \] \[np=a+pn-p\,\,\,\,\Rightarrow \,\,\,a=p\] \[\therefore \]\[{{S}_{P}}=\frac{p}{2}[2p+(p-1)\times 2p]=p[p+{{p}^{2}}-p]={{p}^{3}}\]
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