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Ex 9.3, 23 - If first and nth term of GP are a and b - Geometric Progression(GP): Calculation based/Proofs

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Ex 9.3, 23 If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n. Let a be the first term of G.P & r be the common ratio of G.P Given, first term of G.P = a We know that nth term of G.P = arn-1 b = arn-1 Now P is the product of n terms P = a1 × a2 × a3 × … an = a × ar × ar2 × ar3 … × arn – 1 = (a × a × … a) × (r × r2 × …rn–1) = an 𝑟^(1+2+…+(𝑛−1)) = an 𝑟^((𝑛(𝑛 − 1))/2) Thus, P = an 𝑟^((𝑛(𝑛 − 1))/2) We need to prove P2 = (ab)n. Taking L.H.S P2 = ("an " r^((n(n − 1))/2) )^2 = (𝑎^(𝑛 × 2) " " r^((n(n − 1))/2 × 2) ) = (𝑎^2𝑛 " " r^(n(n − 1)) ) = (𝑎^2 " " r^((n − 1)) )^𝑛 = (𝑎 × 𝑎 × r^((n − 1)) )^𝑛 = (𝑎 × (𝑎r^((n − 1) )))^𝑛 = (𝑎 ×𝑏)^𝑛 = (𝑎𝑏)^𝑛 = R.H.S Thus, P2 = (𝑎𝑏)^𝑛 Hence proved

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