Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg

Last updated at March 9, 2017 by Teachoo

Transcript

Ex 9.3, 3 The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps. We know that an = arn – 1 where an = nth term of GP n is the number of terms a is the first term r is the common ratio Here, 5th term is p i.e. a5 = p Putting n = 5 in an formula p = ar5–1 p = ar4 Also, 8th term is q i.e. a8 = q Putting n = 8 in an formula q = ar8 – 1 q = ar7 Also, 11th term is s i.e. a11 = s Putting n = 11 in an formula s = ar11 – 1 s = ar10 We need to show that q2 = ps. Taking L.H.S q2 Putting value q = ar7 from (2) = (ar7)2 = a2r14 Now, solving R.H.S ps Putting p = ar4 , s = ar10 from (1)& (3) = (ar4) (ar10) = (a × a)(r10 × r4) = a2r14 = L.H.S Thus, L.H.S = R.H.S Hence proved

Ex 9.3

Ex 9.3, 1

Ex 9.3, 2

Ex 9.3, 3 Important You are here

Ex 9.3, 4

Ex 9.3, 5

Ex 9.3, 6

Ex 9.3, 7

Ex 9.3, 8

Ex 9.3, 9

Ex 9.3, 10

Ex 9.3, 11 Important

Ex 9.3, 12

Ex 9.3, 13

Ex 9.3, 14

Ex 9.3, 15

Ex 9.3, 16

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

Ex 9.3, 24

Ex 9.3, 25

Ex 9.3, 26

Ex 9.3, 27 Important

Ex 9.3, 28 Important

Ex 9.3, 29 Important

Ex 9.3, 30

Ex 9.3, 31

Ex 9.3, 32

Chapter 9 Class 11 Sequences and Series

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.