Ex 9.3, 9 - Chapter 9 Class 11 Sequences and Series

Last updated at May 29, 2018 by Teachoo

Ex 9.3, 9 - Find sum to n terms in GP 1, -a, a2, -a3 - Geometric Progression(GP): Formulae based

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)

Ex 9.3

Ex 9.3, 9 You are here

Ex 9.4 →

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.