Question

there are n A. M. 's between 7and 85 such that (n-3)th mean : nth mean is as 11 : 24; find n. pls answer quickly

Answer 1

 are inserted between 7 and 85

So, 7, a1 , a2 , a3 ,.....................am , 85 are in AP

Total number of terms = m + 2

Now first term a = 7

Now Tm+2 = a + (m + 2 -1)d

=> 85 = 7 + (m + 1)d

=> 85 - 7 = (m + 1)d

=> 78 = (m + 1)d

=> d = 78/(m + 1)

Now, given

      (m –3)th : m th = 11 : 24

=> (m –3)th / m th = 11 / 24

=> Tm-3 = /Tm = 11/24

=> {a + (m - 3 -1)d}/{a + (m - 1)d} = 11/24

=> {a + (m - 4)d}/{a + (m - 1)d} = 11/24

=> 24{7 + (m - 4)d} = 11{7 + (m - 1)d}

=> 168 + 24md - 96d = 77 + 11md - 11d

=>  168 + 24md  - 77 - 11md = 96d - 11d

=> 91 + 13md = 85d

=> 91 = 85d - 13md

=> 91 = (85 - 13m)d

=> 91 = (85 - 13m)*{78/(m + 1)}

=> 7 = (85 - 13m)*{6/(m + 1)}

=> 7(m + 1) = 6(85 - 13m)

=> 7m + 7 = 510 - 78m

=> 7m + 78m = 510 - 7

=> 85m = 503

=> m = 503/85

=> m = 5.82

Since m is not an integer. So, there is something wrong in the question.