Ex 9.3
Ex 9.3, 2
Ex 9.3, 3 Important
Ex 9.3, 4
Ex 9.3, 5 (a)
Ex 9.3, 5 (b) Important
Ex 9.3, 5 (c)
Ex 9.3, 6
Ex 9.3, 7 Important
Ex 9.3, 8
Ex 9.3, 9 Important
Ex 9.3, 10
Ex 9.3, 11 Important
Ex 9.3, 12
Ex 9.3, 13
Ex 9.3, 14 Important
Ex 9.3, 15
Ex 9.3, 16 Important
Ex 9.3, 17 Important
Ex 9.3, 18 Important
Ex 9.3, 19
Ex 9.3, 20
Ex 9.3, 21
Ex 9.3, 22 Important
Ex 9.3, 23 Important You are here
Ex 9.3, 24
Ex 9.3, 25
Ex 9.3, 26 Important
Ex 9.3, 27 Important
Ex 9.3, 28
Ex 9.3, 29 Important
Ex 9.3, 30 Important
Ex 9.3, 31
Ex 9.3, 32 Important
Ex 9.3
Last updated at Sept. 3, 2021 by Teachoo
Ex 9.3, 23 If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n. Let a be the first term of G.P & r be the common ratio of G.P Given, first term of G.P = a We know that nth term of G.P = arn-1 b = arn-1 Now P is the product of n terms P = a1 a2 a3 an = a ar ar2 ar3 arn 1 = (a a a) (r r2 rn 1) = an ^(1+2+ +( 1)) = an ^(( ( 1))/2) Thus, P = an ^(( ( 1))/2) We need to prove P2 = (ab)n. Taking L.H.S P2 = ("an " r^((n(n 1))/2) )^2 = ( ^( 2) " " r^((n(n 1))/2 2) ) = ( ^2 " " r^(n(n 1)) ) = ( ^2 " " r^((n 1)) )^ = ( r^((n 1)) )^ = ( ( r^((n 1) )))^ = ( )^ = ( )^ = R.H.S Thus, P2 = ( )^ Hence proved