Ex 8.2, 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)

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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.