Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex 9.3, 9 - Chapter 9 Class 11 Sequences and Series (Term 1)
Last updated at Sept. 3, 2021 by Teachoo
Ex 9.3
Ex 9.3, 1
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 You are here
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
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
Chapter 9 Class 11 Sequences and Series
Serial order wise
- Ex 9.1
- Ex 9.2
-
Ex 9.3
- Ex 9.4
- Examples
- Miscellaneous
Transcript
Ex 9.3, 9 Find the sum to n terms in the geometric progression 1, –a, a2, – a3….. (if a ≠ -1) 1, –a, a2, – a3….. Since small ‘a’ is used here, we used ‘A’ for first term We know that Sn = (A(1 − R^𝑛))/(1 − R) where Sn = sum of n terms of GP n is the number of terms A is the first term R is the common ratio First term = A = 1 Common ratio = R = (−𝑎)/1 Now, sum of n terms = (A(1 − R^𝑛))/(1 − R) Putting values A = 1, R = – a Sn = (1[1−(−a)^𝑛])/(1−(−a)) = ([1−(−a)^𝑛])/(1 + a) Hence sum of n terms is ([1−(−a)^𝑛])/(1 + a)
