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If it is an AP,Sn=Pn2 and Sm =Pm2 m is not equal to n in an A.P, where Sr denotes the sum of r terms of A.P. find Sp

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S_n = P n^2 =\frac{1}{2}n[ 2a + (n-1)d]\\\\2P\ n=2a+(n-1)d,\ \ -- eq2\\\\similarly,\ 2Pm=2a+(m-1)d,\ \ -- eq1\\\\Subtract\ eq1\ from\ eq2,\\\\2P(n-m)=(n-m)d\\\\2P=d,\ \ as\ n \neq m\\\\substitute\ in\ equ1\\\\2Pm=2a+(m-1)2P\\\\2a-2P=0,\ \ a=P\\

AP series :  P + 3P + 5P + 7P + ....

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