Question

Between 1 and 31 are inserted 'm' arithmetic means, so that the ratio of 7th and [m-1]th means is 5:9 then find m ?

Answer 1

K th Arithmetic mean of a sequence of A.P $(A_K)$ = $a+ k\frac{b-a}{n+1}$ , Where n is total number of A.M's b/w a and b .
Here,  
           a = 1 , b = 31  and n = m
A/Q,
            7th  A.M : (m-1)th A.M = 5:9
or,    $\frac{1+ \frac{7(31-1)}{m+1} }{1+ \frac{(m-1)(31-1)}{m+1} } = \frac{5}{9}$
 or,    $\frac{m+1+7*30}{m+1+30(m-1)} = 5/9$
 
 or,    $\frac{m+211}{31m-29} =5/9$

or,       $9m+1899 = 155m-145$

or,     $146m = 2044$

therefore $m = \frac{2044}{146} = 14$

Answer 2

In an arithmetic series, we have n terms, or m+2 terms as in : 
$a_0, a_1,a_2.....,a_{m-1}, a_{m}, a_{m+1}\\ a_0=1,\ \ \ a_{m+1}=31$

We have m = (n - 2) arithmetic means.      The common difference is 
$d =\frac{(a_{n-1} - a_0 )}{n-1}=\frac{a_{m+1}- a_0}{m+1}\\d=\frac{31-1}{m+1}=\frac{30}{m+1}$

$7th\ mean = a_7 = a_0 + 7 d=1+\frac{210}{m+1}\\ \\(m-1)th\ mean=a_0 + (m-1) d=1+\frac{30(m-1)}{m+1}\\ \\Ratio=\frac{m+1+210}{m+1+30(m-1)}=\frac{5}{9}$

$9m+1899=5*31m-29*5\\ \\146 m = 1899+145\\ \\m=2044/146 = 14$

d = 30/15 = 2
The arithmetic series is 1, 3, 5, 7 ,...  27, 29, 31
There are totally 16 terms in this series.   There are 14 arithmetic means in between 1 and 31.