# Misc 3 - 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 3 Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively . Show that S3 = 3 (S2– S1) We know that Sum of n terms = n/2 [2a + (n – 1)d] i.e. Sn = n/2 [2a + (n – 1)d] where a is the first term & d is the common difference of the A.P. It is given that Sum of n terms = S1 i.e. n/2 [2a+(n – 1)d] = S1 Also, Sum of 2n terms = S2 i.e. 2n/2 [2a + (2n – 1)d] = S2 And Sum of 3n terms = S3 i.e. 3n/2 [2a + (3n – 1)d] = S3 We want to show S3 = 3(S2 – S1) Taking R.H.S 3(S2 – S1) = 3[2n/2 " [2a + (2n" −"1)d]" −n/2 " [2a + (n" −"1)d]" ] = 3 (n/2)["2[2a + (2n" −"1)d]" −"[2a + (n" −"1)d]" ] = 3n/2 ["2(2a) + 2(2n" −"1)d" −"2a" −"(n" −"1)d" ] = 3n/2 ["4a + (4n" −"2)d" −"2a" −"(n" −"1)d" ] = 3n/2 ["4a + 4nd" −"2d" −"2a" −"nd + d" ] = 3n/2 ["4a" −"2a + 4nd" −"nd" −"2d + d" ] = 3n/2 ["2a + 3nd" −"d" ] = 3n/2 ["2a + (3n – 1)d" ] From (3), S3 = 3𝑛/2 [2a + (3n – 1)d = S3 = L.H.S Hence, R.H.S = L.H.S Hence proved

Miscellaneous

Misc 1

Misc 2

Misc 3 Important You are here

Misc 4

Misc 5

Misc 6

Misc 7 Important

Misc 8

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 9 years. He provides courses for Maths and Science at Teachoo.