Ex 8.2, 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)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.