Example 12 - Chapter 9 Class 11 Sequences and Series (Term 1)
Last updated at Sept. 3, 2021 by Teachoo
Examples
Example 1 (i)
Example 1 (ii)
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Example 3 Important
Example 4
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Example 7
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Example 9
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Example 11
Example 12 Important You are here
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Example 15 Important
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Example 18 Important
Example 19 Important Deleted for CBSE Board 2022 Exams
Example 20 Deleted for CBSE Board 2022 Exams
Example 21 Important
Example 22
Example 23
Example 24 Important
Chapter 9 Class 11 Sequences and Series (Term 1)
Serial order wise
- Ex 9.1
- Ex 9.2
- Ex 9.3
- Ex 9.4
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- Miscellaneous
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
