Question:
Between 7 and 85, m number of arithmetic means are inserted so that the ratio of (m –3)th and m th means is 11 : 24. Find the value of m.
Answer:
Given there are m number of arithmetic means 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.
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