web analytics

Example 15 - Find sum of 7, 77, 777, 7777, ... to n terms - Geometric Progression(GP): Formulae based

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
Ask Download

Transcript

Example 15 Find the sum of the sequence 7, 77, 777, 7777, ... to n terms. 7, 77, 777, 7777, ... n terms Here, 77/7 = 11 & 777/77 = 10.09 Thus, (π‘†π‘’π‘π‘œπ‘›π‘‘ π‘‘π‘’π‘Ÿπ‘š)/(πΉπ‘–π‘Ÿπ‘ π‘‘ π‘‘π‘’π‘Ÿπ‘š) β‰  (π‘‡β„Žπ‘–π‘Ÿπ‘‘ π‘‘π‘’π‘Ÿπ‘š)/(π‘†π‘’π‘π‘œπ‘›π‘‘ π‘‘π‘’π‘Ÿπ‘š) i.e. common ratio is not same ∴ This is not a GP We need to find sum Sum = 7 + 77 + 777 + 7777 + ...upto n terms = 7(1 + 11 + 111 + …. … upto n terms) = 7(1 + 11 + 111 + …. … upto n terms) Multiplying & dividing by 9 = 7/9 [9(1 + 11 + 111 + …upto n term) = 7/9 [9 + 99 + 999 + 9999 + …upto n terms] = 7/9 [(10 – 1) + (100 – 1) + (1000 – 1) +…upto n terms] = 7/9 [(10 + 100 + 1000 + ….n terms) – 1 – 1 – 1 –…upto n terms] = 7/9 [(10 + 100 + 1000 + ….n terms) – (1 + 1 + 1 +…upto n terms)] = 7/9 [(10 + 100 + 1000 + ….n terms) – n Γ— 1] = 7/9 [(10 + 100 + 1000 + ….n terms) – n] Now, a = 10, r = 10 For, r > 1 i.e. Sn = (a(π‘Ÿ^π‘›βˆ’ 1))/(π‘Ÿ βˆ’ 1) Putting value of a = 10 & r = 10 Sn = (10(γ€–10γ€—^𝑛 βˆ’ 1))/(10 βˆ’ 1) Sn = (10(γ€–10γ€—^𝑛 βˆ’ 1))/9 Now substituting this value in (1) Sum = 7/9 [(10 + 102 + 103 + … upto n terms) – n] Sum = 7/9 [(10(γ€–10γ€—^𝑛 βˆ’ 1))/9 " – n" ] Thus, 7, 77, 777, 7777, ...upto n terms = 7/9 [(10(γ€–10γ€—^𝑛 βˆ’ 1))/9 " – n" ]

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail