If the sum of p terms of an arithmetic progre | vedclass

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(C) Given that the sum of $p$ terms equals the sum of $q$ terms,we have:$\frac{p}{2}[2a + (p - 1)d] = \frac{q}{2}[2a + (q - 1)d]$Expanding both sides:

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$0$

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(C) Given that the sum of $p$ terms equals the sum of $q$ terms,we have:$\frac{p}{2}[2a + (p - 1)d] = \frac{q}{2}[2a + (q - 1)d]$Expanding both sides:$p[2a + (p - 1)d] = q[2a + (q - 1)d]$$2ap + p(p - 1)d = 2aq + q(q - 1)d$Rearranging the terms:$2a(p - q) + d[p(p - 1) - q(q - 1)] = 0$$2a(p - q) + d[p^2 - p - q^2 + q] = 0$$2a(p - q) + d[(p^2 - q^2) - (p - q)] = 0$Factoring out $(p - q)$:$(p - q)[2a + d(p + q - 1)] = 0$Since $p 
eq q$,we must have:$2a + (p + q - 1)d = 0$The sum of $(p + q)$ terms is given by:$S_{p+q} = \frac{p+q}{2}[2a + (p + q - 1)d]$Substituting the value $2a + (p + q - 1)d = 0$:$S_{p+q} = \frac{p+q}{2} 	imes 0 = 0$

If the sum of $p$ terms of an arithmetic progression is equal to the sum of its $q$ terms,then what is the sum of its $(p + q)$ terms?

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