1. Class 11
2. Important Question for exams Class 11

Transcript

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 ๐ก๐๐๐ ๐๐ 1๐ ๐ก ๐ด๐ )/(18๐กโ ๐ก๐๐๐ ๐๐ 2๐๐ ๐ด๐) = (๐18 ๐๐ 1๐ ๐ก ๐ด๐ )/(๐ด18 ๐๐ 2๐๐ ๐ด๐) = (a + (18 โ 1)d)/(A + (18 โ 1)D) = (๐ + 17๐)/(A + 17D) is given that (Sum of n terms of first A๐)/(Sum of n terms of second A๐) = (5n+4)/(9n+6) (๐/2[2๐+(๐ โ 1)๐])/((๐ )/2[2๐ด+(๐ โ 1)๐ท]) = (5n+4)/(9n+6) ( [2๐+(๐ โ 1)๐])/( [2๐ด+(๐ โ 1)๐ท]) = (5n+4)/(9n+6) ( 2(a +((๐ โ1)/2)d))/( 2(A +((๐ โ1)/2)D) ) = (5n+4)/(9n+6) ( (a +((๐ โ1)/2)d))/( (A +((๐ โ1)/2)D) ) = (5n+4)/(9n+6) We have to find (๐ + 17๐)/(A + 17D) Hence, (๐ โ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 โ1)/2)d))/( (A +((35 โ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 ๐ก๐๐๐ ๐๐ 1๐ ๐ก ๐ด๐ )/(18๐กโ ๐ก๐๐๐ ๐๐ 2๐๐ ๐ด๐) = 179/321 Hence the ratio of 18th term of 1st AP and 18th term of 2nd AP is 179 : 321

Class 11
Important Question for exams Class 11