Ex 9.2, 9 - Sums of n terms of two APs are in ratio 5n+4: 9n+6 - Arithmetic Progression (AP): Formulae based

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Ex 9.2 , 9 The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms. There are two AP with different first term and common difference For the first AP Let first term be = a Common difference = d Sum of n terms = Sn = ๐‘›/2 [2a + (n โ€“ 1)d] & nth term = an = a + (n โ€“ 1)d For the second AP Let first term be = A common difference = D Sum of n terms = Sn = ๐‘›/2 [2A + (n โ€“ 1)D] & nth term = An = A + (n โ€“ 1)D We need to find ratio of their 18th term i.e. (18๐‘กโ„Ž ๐‘ก๐‘’๐‘Ÿ๐‘š ๐‘œ๐‘“ 1๐‘ ๐‘ก ๐ด๐‘ƒ )/(18๐‘กโ„Ž ๐‘ก๐‘’๐‘Ÿ๐‘š ๐‘œ๐‘“ 2๐‘›๐‘‘ ๐ด๐‘ƒ) = (๐‘Ž18 ๐‘œ๐‘“ 1๐‘ ๐‘ก ๐ด๐‘ƒ )/(๐ด18 ๐‘œ๐‘“ 2๐‘›๐‘‘ ๐ด๐‘ƒ) = (a + (18 โˆ’ 1)d)/(A + (18 โˆ’ 1)D) = (๐‘Ž + 17๐‘‘)/(A + 17D) is given that (Sum of n terms of first A๐‘ƒ)/(Sum of n terms of second A๐‘ƒ) = (5n+4)/(9n+6) (๐‘›/2[2๐‘Ž+(๐‘› โˆ’ 1)๐‘‘])/((๐‘› )/2[2๐ด+(๐‘› โˆ’ 1)๐ท]) = (5n+4)/(9n+6) ( [2๐‘Ž+(๐‘› โˆ’ 1)๐‘‘])/( [2๐ด+(๐‘› โˆ’ 1)๐ท]) = (5n+4)/(9n+6) ( 2(a +((๐‘› โˆ’1)/2)d))/( 2(A +((๐‘› โˆ’1)/2)D) ) = (5n+4)/(9n+6) ( (a +((๐‘› โˆ’1)/2)d))/( (A +((๐‘› โˆ’1)/2)D) ) = (5n+4)/(9n+6) We have to find (๐‘Ž + 17๐‘‘)/(A + 17D) Hence, (๐‘› โˆ’1)/2 = 17 n โ€“ 1 = 17 ร— 2 n โ€“ 1 = 34 n = 34 + 1 n = 35 Putting n = 35 in (1) ( (a +((35 โˆ’1)/2)d))/( (A +((35 โˆ’1)/2)D) ) " "= (5(35)+4)/(9(35)+6) ( (a +(34/2)d))/( (A +(34/2)D) )= (175 + 4)/(315 + 6) ([a + 17d])/( [A + 17D]) = 179/321 Therefore (18๐‘กโ„Ž ๐‘ก๐‘’๐‘Ÿ๐‘š ๐‘œ๐‘“ 1๐‘ ๐‘ก ๐ด๐‘ƒ )/(18๐‘กโ„Ž ๐‘ก๐‘’๐‘Ÿ๐‘š ๐‘œ๐‘“ 2๐‘›๐‘‘ ๐ด๐‘ƒ) = 179/321 Hence the ratio of 18th term of 1st AP and 18th term of 2nd AP is 179 : 321

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