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Ex 9.3, 18 - Find sum to n terms of sequence 8, 88, 888, 8888 - Geometric Progression(GP): Formulae based

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Ex9.3, 18 Find the sum to n terms of the sequence, 8, 88, 888, 8888… 8, 88, 888, 8888… to n term This is not a GP but we can relate it to a GP By writing as Sum = 8 + 88 + 888 + 8888 + … upto n terms = 8(1) + 8(11) + 8(111) + … upto n term Taking 8 common = 8(1 + 11 + 111 + … upto n term) Divide & multiply by 9 = 8/9[9(1 + 11 + 111 + … upto n term)] = 8/9 [ 9 + 99 + 999 + 9999 +…upto n terms] Sum = 8/9 [ 9 + 99 + 999 + 9999 +…upto n terms] = 8/9 [ (10 – 1)+(100 – 1)+(1000 – 1)+ …upto n terms] = 8/9 [ (10 – 1)+(102 – 1)+(103 – 1)+ …upto n terms] = 8/9 [ (10 + 102 + 103 … upto n terms) – (1 + 1 + 1 +…upto n terms)] = 8/9 [(10 + 102 + 103 … upto n terms) – n × 1] We will solve (10 + 102 + 103 … upto n terms) separately We can observe that this is GP With first term a = 10 & common ratio r = 102/10 = 10 We know that sum of n terms = (a(𝑟^𝑛− 1))/(𝑟 − 1) i.e. Sn =(a(𝑟^𝑛− 1))/(𝑟 − 1) putting value of a & r Sn = (10(10n − 1))/(10 − 1) Substituting 10 + 102 + 103 +…upto n times = (10(10n − 1))/9 in (1) Sum = 8/9 [(10 + 102 + 103 +…upto n terms) – n] = 8/9 [ (10(10n−1))/(10−1) – n ] = 8/9 [ (10(10n−1))/9 – n ] = 8/9 [ (10(10n−1) − 9n)/9] = 8/9 [ (10(10n−1) − 9n)/9] = 8/9 [ 10(10n−1)/9 – 9/9n] = 8/9 × 10(10n−1)/9 – 8/9 × 9/9n] = 80/81(10n – 1) - 8/9n Hence sum of sequence 8 + 88 + 888 + 8888 + …………….. to n terms = = 80/81(10n – 1) – 8/9n

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