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
Answers
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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