# Misc 8 - Chapter 9 Class 11 Sequences and Series

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 8 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms. Let a be the first term of GP & r be the common ratio of GP It is given that a = 5 & r = 2 (r > 1) Sum of some term of a GP = 315 Let the sum of n terms of GP = 315 Sn = 315 We know that Sum of n terms of GP = (a( ^ 1))/(r 1) Sn = (a( ^ 1))/(r 1) 315 = (a( ^ 1))/(r 1) Putting r = 2 & a = 5 315 = (5(2n 1))/(2 1) 315 = (5(2n 1))/1 315 = 5 (2n 1) 315/5 = 2n 1 63 = 2n 1 64 = 2n (2)6 = 2n comparing powers n = 6 Hence number of terms is 6 We need to find last term of GP Let l be the last term of GP We know that n term of GP an = arn 1 l = arn 1 Putting a = 5 , r = 2 & n = 6 l = 5(2)6 1 = 5(2)5 = 5(2 2 2 2 2) = 5 64 = 320 Hence, the last term is 320

Miscellaneous

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6

Misc 7 Important

Misc 8 You are here

Misc 9

Misc 10

Misc 11

Misc 12

Misc 13

Misc 14

Misc 15

Misc 16 Important

Misc 17

Misc 18

Misc 19 Important

Misc 20

Misc 21

Misc 22

Misc 23

Misc 24

Misc 25 Important

Misc 26

Misc 27

Misc 28 Important

Misc 29

Misc 30

Misc 31

Misc 32 Important

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.