Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex 9.3, 10 Find the sum to n terms in the geometric progression x3, x5, x7….. (if x ≠ ± 1) x3, x5, x7….. 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 Here, First term a = x3 Common ratio r = 𝑥5/𝑥3 = x2 Now, ∴ Sum of n terms = (a[1 − 𝑟^𝑛])/(1 − r) Putting values a = x3 & r = x2 Sn = (x3[1 − (x2)n])/(1 − x2) = (x3(1 − x2n))/(1 − x2) Hence sum of n terms is (𝑥3 [ 1 − 𝑥2𝑛])/(1 −𝑥2)
Ex 9.3
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
Ex 9.3, 10 You are here
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
Ex 9.3
About the Author