Answers

2014-10-09T17:05:26+05:30
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
0
2044/146 = 14 always not in fraction.
yeah i think i do this type of sum by taking the upper value
karthik u r answer is absolutely correct.
good. right answer
ok i got my mistake
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2014-10-09T17:48:54+05:30

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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.


2 4 2
wait a bit please.
its done.
thanx n u r welcom