If the sum of the first p terms of an A.P. is | vedclass

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(A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]

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$0$

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(A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Given that $S_p = S_q$,we have:$\frac{p}{2}[2a + (p-1)d] = \frac{q}{2}[2a + (q-1)d]$Multiplying both sides by $2$:$p[2a + (p-1)d] = q[2a + (q-1)d]$$2ap + p(p-1)d = 2aq + q(q-1)d$Rearranging the terms:$2a(p-q) + [p(p-1) - q(q-1)]d = 0$$2a(p-q) + [p^2 - p - q^2 + q]d = 0$$2a(p-q) + [(p^2 - q^2) - (p-q)]d = 0$$2a(p-q) + [(p-q)(p+q) - (p-q)]d = 0$Dividing by $(p-q)$ (assuming $p 
eq q$):$2a + (p+q-1)d = 0$Now,the sum of the first $(p+q)$ terms is:$S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d]$Substituting $2a + (p+q-1)d = 0$:$S_{p+q} = \frac{p+q}{2} 	imes 0 = 0$Thus,the sum of the first $(p+q)$ terms is $0$.

If the sum of the first $p$ terms of an $A.P.$ is equal to the sum of the first $q$ terms,then find the sum of the first $(p+q)$ terms.

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