Question

There are n arithmetic means between 20 and 80 such that first mean : last mean=1:3, find n.

Answer 1

Let A1, A2, A3, …….An be n arithmetic means between 20 and 80 and let d be the common difference between the terms of the A.P.

Then, d = (b – a)/(n + 1)
              = (80 – 20)/(n + 1)
              = 60 / (n + 1)

A1 = 20 + d
      = 20 + 60 / (n + 1)
      = 20[(n + 4)/(n + 1)]

An = 20 + d
      = 20 + 60n / (n + 1)
      = 20 [(4n + 1) / (n + 1)]

A1/An = 1/3

⇒ 20[(n + 4)/(n + 1)] / 20 [(4n + 1) / (n + 1)] = 1/3 ⇒ (n + 4) / (4n + 1) = 1/3 ⇒ 4n + 1 = 3n + 12 ⇒ n = 11.

Answer 2

Given : n arithmetic means are inserted between 20 and 780 . ratio  of the first and last arithmetic mean is 1:3

To find : Value of n

Solution:

if n arithmetic means are inserted between 20 and 80

Then Series becomes

20  , 20 + d  ,  20 + 2d  + ................................. , 20 + nd  , 80

total Terms = n + 2

970 = 20 + (n + 2 - 1) d = 20 + nd  + d

=> 20 + nd = 80 - d

ratio of first and last arithmetic mean   is 1 :3

(20 + d)/(20 + nd)  = 1/3

=> (20 + d)/(80 - d)  = 1/3

=> 60 + 3d  =80 - d

=> 4d = 120

=> d = 5

20 + nd = 80 - d

=> 20 +  n(5 ) = 7 60  - 5

=> n(5)  = 55

=> n  = 11

11 arithmetic means are inserted between 20 and 70  

Series becomes

20 , 25  , 30 , ........................, 75 , 80

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