Question

if n arithmetic means are inserted between 20 and 70 and the ration of the first and last arithmetic mean is 1:3 then n​

Answer 1

Given : n arithmetic means are inserted between 20 and 70 . ration  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 70

Then Series becomes

20  , 20 + d  ,  20 + 2d  + ................................. , 20 + nd  ,  70

total Terms = n + 2

70 = 20 + (n + 2 - 1) d = 20 + nd  + d

=> 20 + nd = 70 - d

ratio of first and last arithmetic mean   is 1 :3

(20 + d)/(20 + nd)  = 1/3

=> (20 + d)/(70 - d)  = 1/3

=> 60 + 3d  = 70 - d

=> 4d = 10

=> d = 5/2

20 + nd = 70 - d

=> 20 +  n(5/2) = 70  - 5/2

=> n(5/2)  = 95/2

=> n  = 19

19 arithmetic means are inserted between 20 and 70  

Series becomes

20 , 22.5  , 25  , ........................, 67.5 , 70

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