Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9 .3, 14 The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P. Given Now, Sn = (๐(1 โ ๐^๐))/(1 โ ๐ ) where Sn = sum of n terms of GP n is the number of terms a is the first term r is the common ratio We know that S3 = 16 Putting n = 3 in Sn (๐(1 โ ๐3 ))/(1 โ ๐ ) = 16 Also, S6 = 144 Putting n = 6 in Sn (a(1 โ r6))/(1โ r) = 144 We need to find first term (a) & common ratio (r) Dividing (2) from (1) ((๐(1 โ ๐6))/(1โ ๐))/((๐(1 โ ๐3))/(1 โ ๐)) = 144/16 (๐(1 โ ๐6))/(1โ ๐) ร (1โ ๐)/(๐(1 โ ๐3)) = 144/16 (๐(1 โ ๐6)(1โ ๐))/(๐(1 โ ๐3)(1โ ๐)) = 9/1 (1 โ๐6)/(1 โ๐3) = 9/1 (12 โ(๐3)2)/(1 โ ๐3) = 9/1 Using a2 โ b2 = (a โ b)(a + b) ((1 โ ๐3)(1 + ๐3))/((1 โ ๐3)) = 9/1 ((1+ ๐3))/1 = 9/1 ((1+ ๐3))/1 = 9/1 1 + r3 = 9 r3 = 9 โ 1 r3 = 8 r =โ8 r = โ(2 ร2 ร2) r = 2 Hence common ratio r = 2 Putting r = 2 in (1) (๐(1 โ๐3))/(1 โ๐) = 16 (๐(1 โ23))/(1 โ2) = 16 (๐(1 โ8))/(โ1) = 16 (๐(1 โ8))/(โ1) = 16 (๐(โ7))/(โ1) = 16 a(7) = 16 a = 16/7 Hence first term (a) =16/7 Now, a = 16/7 , r = 16/7 Now we need to find sum of n terms As r > 1 Sum of n terms of G.P = (๐(๐^๐ โ 1))/(๐ โ 1) Sum of n terms of G.P = (๐(๐^๐ โ 1))/(๐ โ 1) Putting a = 16/7 & r = 2 Sn = (16/7(2๐ โ 1 ))/(2 โ 1 ) = (16/7(2๐ โ 1 ))/1 = 16/7 (2n โ 1) Hence sum of n terms is 16/7 (2n โ 1)

Ex 9.3

Ex 9.3, 1

Ex 9.3, 2

Ex 9.3, 3 Important

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 You are here

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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.