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Ex 9.3, 24 - Show ratio of sum of n terms of GP, sum from - Geometric Progression(GP): Calculation based/Proofs

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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We need to show ratio of the sum of n terms of GP & sum of terms from (n + 1)th to 2nth term i.e. we need to calculate (๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘› ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘ )/(๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘  ๐‘“๐‘Ÿ๐‘œ๐‘š (๐‘› + 1)๐‘กโ„Ž ๐‘ก๐‘œ (2๐‘›)๐‘กโ„Ž ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘ ) Putting values from (1) & (2) = (( a(1 โˆ’ใ€– ๐‘Ÿใ€—^๐‘›))/((1 โˆ’ r)))/(( a)/((1 โˆ’ r))(ใ€– ๐‘Ÿใ€—^๐‘›โˆ’ใ€– ๐‘Ÿใ€—^2๐‘›)" " ) = (( a(1 โˆ’ใ€– ๐‘Ÿใ€—^๐‘›))/((1 โˆ’ r)))/(( a)/((1 โˆ’ r) ) ใ€– ๐‘Ÿใ€—^๐‘› (1 โˆ’ใ€– ๐‘Ÿใ€—^๐‘›)" " ) = (( a(1 โˆ’ใ€– ๐‘Ÿใ€—^๐‘›))/((1 โˆ’ r)))/๐‘Ÿ๐‘›(( a(1 โˆ’ใ€– ๐‘Ÿใ€—^๐‘›))/((1 โˆ’ r))) = ( a(1 โˆ’ ๐‘Ÿ^๐‘›))/(rn(1 โˆ’ใ€– ๐‘Ÿใ€—^๐‘›)) ร— ( 1 โˆ’ r)/(1 โˆ’ r) Ex 9.3, 24 Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n+1)th to (2n)th term is ( 1)/rn Ex 9.3, 24 Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n+1)th to (2n)th term is ( 1)/rn We need to show ratio of the sum of n terms of GP & sum of terms from (n + 1)th to 2nth term i.e. we need to calculate (๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘› ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘ )/(๐‘†๐‘ข๐‘š ๐‘œ๐‘“ ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘  ๐‘“๐‘Ÿ๐‘œ๐‘š (๐‘› + 1)๐‘กโ„Ž ๐‘ก๐‘œ (2๐‘›)๐‘กโ„Ž ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘ ) Putting values from (1) & (2) = (( a(1 โˆ’ใ€– ๐‘Ÿใ€—^๐‘›))/((1 โˆ’ r)))/(( a)/((1 โˆ’ r))(ใ€– ๐‘Ÿใ€—^๐‘›โˆ’ใ€– ๐‘Ÿใ€—^2๐‘›)" " ) = (( a(1 โˆ’ใ€– ๐‘Ÿใ€—^๐‘›))/((1 โˆ’ r)))/(( a)/((1 โˆ’ r) ) ใ€– ๐‘Ÿใ€—^๐‘› (1 โˆ’ใ€– ๐‘Ÿใ€—^๐‘›)" " ) = (( a(1 โˆ’ใ€– ๐‘Ÿใ€—^๐‘›))/((1 โˆ’ r)))/๐‘Ÿ๐‘›(( a(1 โˆ’ใ€– ๐‘Ÿใ€—^๐‘›))/((1 โˆ’ r))) = ( a(1 โˆ’ ๐‘Ÿ^๐‘›))/(rn(1 โˆ’ใ€– ๐‘Ÿใ€—^๐‘›)) ร— ( 1 โˆ’ r)/(1 โˆ’ r) = 1/๐‘Ÿ๐‘›

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