The ratio of first to the last of n A.m's between 5 and 35 is 1:4. The value of n is
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Answer:
We are given that there are n A. M. s between 5 and 35 such that 2nd mean: last mean 1:4.
Let the series be 5, A1,A2,A3.....An , 35 which means there are n A.M.'s between 5 and 35.
Here a = first term = 5 and let d = common difference.
So, 2nd mean = 3rd term of the series
A3 = a + (n - 1)d = a + 2d
Simiarly, last mean = (n+1)th term of the series
An+1 = a + (n - 1)d = a + (n + 1 - 1)d = a + nd
Now, the ratio given to us is;
a+2d/a+nd=1/4
4a + 8d = a + nd
nd = 3a + 8d
nd = 15 + 8d {as given a = 5} ------------ [Equation 1]
Now, as we know that if there are n arithmetic means between two number, then there are (n+2) terms in an A.P.
So, = a + (n - 1)d = a + (n + 2 - 1)d = a + (n + 1)d
35 = 5 + (n + 1)d
30 = nd + d
nd = 30 - d
Now, putting value of nd in equation 1 we get;
30 - d = 15 + 8d
8d + d = 30 - 15
d = 15/9
Put this value of d in equation 1 we get;
nd = 15 + 8d
n = 15+8d/d
n = 15+(8*15/9)
n = 135/120/9*9/15=225/15
n = 17
The value of n is 17.
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Given : The ratio of first to the last of n A.M's between 5 and 35 is 1:4.
To Find : The value of n
Solution:
First number = 5
Last number = 35
Inserted n A.M's
Hence total numbers = n + 2
a = 5
n + 2 th Term = 35
n + 2 th term = a + (n +2 - 1)d
=> 35 = 5 + (n + 1)d
=> 30 = (n + 1)d
First AM = 5 + d
Last AM = 35 - d
The ratio of first to the last is 1: 4
=> (5 + d)/(35 - d) = 1/4
=> 20 + 4d = 35 - d
=> 5d = 15
=> d = 3
30 = (n + 1)d
=> 30 = (n + 1)3
=> 10 = n + 1
=> n = 9
The value of n is 9
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