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