Example 12 - Chapter 9 Class 11 Sequences and Series
Last updated at May 29, 2018 by Teachoo
Transcript
Example 12 Find the sum of first n terms and the sum of first 5 terms of the geometric series 1 + 2/3 "+" 4/9 + ……… 1 + 2/3 "+" 4/9 + ……… 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 = 1 and common ratio = r = (2/3)/1 Now, we need to calculate sum of n terms of GP We use the formula, Sn = (a(1 − 𝑟^𝑛))/(1 − r) Putting a = 1 , r = 2/3 = 1(1 − ( 2/3)^𝑛 )/(1 − 2/3) = (1 − ( 2/3)^𝑛)/((3 − 2)/3) = (1 − ( 2/3)^𝑛)/(1/3) = 3/1 ["1 –" (2/3)^𝑛 ] = 3["1 –" (2/3)^𝑛 ] Thus, Sum of n terms = Sn = 3["1 –" (2/3)^𝑛 ] Now, we need to find sum of first 5 terms of GP Putting n = 5 in Sn S5 = 3["1 –" (2/3)^5 ] = 3 ["1 –" (2^5/3^5 )] = 3 ["1 –" (32/243)] = 3 [(243 − 32)/243] = 3 [211/243] = 211/81 Hence, sum of first five terms of GP is 211/81
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Chapter 9 Class 11 Sequences and Series
Serial order wise
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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.