Example 13 - How many terms of GP 3, 3/2, 3/4,... are needed - Geometric Progression(GP): Formulae based

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Example 13, How many terms of the G.P. 3, 3/2, 3/4 , , ,... are needed to give the sum 3069/512 ? Here First term = a = 3, Common ratio r = (3/2)/3 = 3/(2 × 3) = 1/2 We know that sum of n term = (𝑎( 1 − 𝑟^𝑛))/(1 −𝑟) Sn = (a(1 − 𝑟^𝑛))/(1−r) Given that Sn = 3069/512 & we need to find n. 3069/512 = (a(1 − 𝑟^𝑛))/(1 − 𝑟) 3069/512 = (3[1 − (1/2)^𝑛])/(1 − 1/2) 3069/512 =(3[1 −(1/2)^𝑛])/( 1/2) 3069/512 = 6[1 – (1/2)^𝑛] 3069/(512 × 6) = 1 – (1/2)^𝑛 3069/3072 = 1 – (1/2)^𝑛 1 – (1/2)^𝑛 = 3069/3072 " " (1/2)^𝑛 " = " ("1 – " 3069/3072) (1/2)^𝑛 = ((3072 − 3069)/3072) (1/2)^𝑛 = (3/3072) (1/2)^𝑛 = (1/1024) (1/2)^𝑛 = (1/2)10 Comparing powers n = 10 Hence 10 terms are needed to give sum 3069/512

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