If the sum of first p terms of an A.P. is equal to the sum of the firs

1.	If the sum of first p terms of an A.P. is equal to the sum of the first q terms, thenfind the sum of the first (p + q) terms.
Answer» <P> $\huge\tt{\bold{\underline{\underline{Question᎓}}}}$ If the sum of 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. $\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$ ┏━━━━━━━━━━━━━━━━━━━━━━━┓ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ✍️ $\bold{\boxed{ Sn = \frac{n}{2} (2a + (n - 1)d}}$ $= > sp = \frac{p}{2} (2a + (p - 1)d$ $= > sq = \frac{q}{2} (2a + (q - 1)d$ $\bold{\red{sp = sq}}$ $= > \frac{p}{2} (2a + (p - 1)d = \frac{q}{2} (2a + (q - 1)d$ $= > p(2a + (p - 1)d = \frac{2 \times q}{2} (2a + (q - 1)d$ $= > p(2a + (p - 1)d) = q(2a + (q - 1)d)$ $= > 2ap + pd(p - 1) = 2aq + qd(q - 1)$ $= > 2ap - 2aq = qd(q - 1) - pd(p - 1)$ $= > 2a(p - q) = d[(q - 1)q - (p - 1)p]$ $= > 2a(p - q) = d( {q}^{2} - q - ( {p}^{2} - p)$ $= > 2a(p - q) = d( {q}^{2} - q - {p}^{2} + p)$ $= > 2a(p - q) = d( {q}^{2} - {p}^{2} + p - q)$ $= > 2a(p - q) = - d( - {q}^{2} + {p}^{2} - p + q)$ $= > 2a(p - q) = - d( {p}^{2} - {q}^{2} - (p - q))$ $= > 2a(p - q) = - d[(p - q)(p + q) - (p - q)]$ $= > 2a(p - q) = - d[(p - q)(p + q) - (p - q) \times 1]$ Taking (p-q) common from R.H.S $\bold{= > (p - q)[2a + d(p + q - 1)d] = 0}$ $= > p - q = 0$ $\bold{ = > p = 0}$ $2a + d(p + q - 1) = 0$ Now applying sum formula: $\bold{ = > sn = \frac{n}{2} [2a + (n - 1)d]}$ $\bold{ = > s(p + q) = \frac{p + q}{2} [2a + (p + q - 1)d]}$ $\bold{= > \frac{p + q}{2} \times 0}$ $= 0$ THEREFORE,the sum of the first (p + q) terms is 0. ┗━━━━━━━━━━━━━━━━━━━━━━━┛