Ex 9.2, 5 - Chapter 9 Class 11 Sequences and Series (Term 1)
Last updated at May 29, 2018 by Teachoo
Last updated at May 29, 2018 by Teachoo
Transcript
Ex 9.2,5 In an A.P., if pth term is 1/q and qth term is 1/p , prove that the sum of first pq terms is 1/2 (pq + 1) where p q. We know that an = a + (n 1)d Where an is nth term of AP, n is the number of terms, a be the first term & d be the common difference of the A.P. It is given that pth term is 1/ i.e. ap = 1/q a + (p 1)d = 1/q Also, qth term of A.P = 1/p i.e. aq = 1/p i.e. a + (q 1)d = 1/p Now subtracting (1) from (2) i.e. (1) (2) [a + (p 1)d] [a + (q 1)d] = 1/ 1/ a + pd d [a + qd d] = 1/ 1/ a + pd d a qd + d = 1/ 1/ a a d + d + pd qd = 1/ 1/ 0 + 0 + pd qd = 1/ 1/ d(p q) = 1/ 1/ d(p q) = ( )/ d = ( )/( ( )) d = 1/ Now finding first term i.e. a Putting d = 1/ in (1) 1/ = a + (p 1)d 1/ = a + (p 1)1/ 1/ = a + / 1/ 1/ = a + 1/ 1/ 1/ 1/ = a 1/ 0 = a 1/ 1/ = a a = 1/ Therefore, a = 1/ Thus, a = 1/ & d = 1/ We need to show sum of first pq term is 1/2(pq + 1) i.e. Spq = 1/2 (pq + 1) 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 For sum of first pq terms, Putting n = pq , a = 1/ , d = 1/ Spq = pq/2 ["2 " 1/ " + (pq 1)" 1/ ] = pq/2 [2/ " + " / 1/ ] = pq/2 [(2 + 1 )/ ] = pq/2 [(1 + )/ ] = 1/2 [pq + 1] Thus, sum of first pq terms is 1/2 (pq+1) Hence proved
Ex 9.2
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