Ex 9.2, 10 - If sum of first p terms of AP is equal to - Arithmetic Progression (AP): Calculation based/Proofs

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Ex9.2 , 10 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. We know that Sn = n/2 ( 2a + (n – 1)d ) Where, Sn = sum of n terms of A.P. n = number of terms a = first term and d = common difference Now, Sum of first p terms = Sp = p/2 [2a + (p – 1)d] Sum of first q terms = Sq = q/2 [2a + (q – 1)d] It is given that Sum of first p terms = Sum of first q terms p/2 [2a + (p – 1)d] = q/2 [2a + (q – 1)d] p[2a + (p – 1)d] = (2 × 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[q2 – q – (p2 – p)] 2a(p – q) = d[q2 – q – p2 + p] 2a(p – q) = d[q2 – p2 + p – q] 2a(p – q) = – d [ –q2 + p2 – p + q] 2a(p – q) = – d [p2 – q2 – (p – q)] 2a(p – q) = – d [(p – q) (p + q) – (p – q)] 2a(p – q) = – d(p – q) [p + q – 1] 2a(p – q) + d(p – q) [p + q – 1] = 0 (p – q) [2a + d(p + q – 1)] = 0 ∴ 2a + d(p + q – 1) = 0 Now, finding sum of first (p + q) terms We know that, Sum of n terms = n/2 [2a + (n – 1)d] For sum of (p + q) terms, we put n = (p + q) Sum of (p + q) term is = (p + q)/2 [2a + (p + q – 1)d] = (p + q)/2 × 0 = 0 Hence, sum of (p + q) term is 0

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