Chapter 9 Class 11 Sequences and Series
Question 5 Important
Question 9 Important
Question 15 Important
Question 17
Example 9 Important
Example 10 Important
Ex 8.2, 3 Important You are here
Ex 8.2, 11 Important
Ex 8.2, 17 Important
Ex 8.2, 18 Important
Ex 8.2, 22 Important
Ex 8.2, 28
Ex 8.2, 29 Important
Ex 9.4.4 Important
Question 7 Important
Question 9 Important
Question 10
Question 9
Question 9 Important
Misc 10 Important
Question 13 Important
Misc 14 Important
Misc 18 Important
Chapter 9 Class 11 Sequences and Series
Last updated at April 16, 2024 by Teachoo
Ex 8.2, 3 The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps. We know that an = arn – 1 where an = nth term of GP n is the number of terms a is the first term r is the common ratio Here, 5th term is p i.e. a5 = p Putting n = 5 in an formula p = ar5–1 p = ar4 Also, 8th term is q i.e. a8 = q Putting n = 8 in an formula q = ar8 – 1 q = ar7 Also, 11th term is s i.e. a11 = s Putting n = 11 in an formula s = ar11 – 1 s = ar10 We need to show that q2 = ps. Taking L.H.S q2 Putting value q = ar7 from (2) = (ar7)2 = a2r14 Now, solving R.H.S ps Putting p = ar4 , s = ar10 from (1)& (3) = (ar4) (ar10) = (a × a)(r10 × r4) = a2r14 = L.H.S Thus, L.H.S = R.H.S Hence proved