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Between 1 and 31, m numbers are inserted in such a way that the ratio of 7th and (m-1)th term in 5:9. Find the value of m.
Let the arithmetic mean inserted be A1, A2, A3,....Am.
now , series is 1,A1, A2, A3,....Am ,31
hence , a = 1 Tn =31 , n=m+2
Tn = a+(n-1)d
31 = 1+(m+2-1)d
30=(m+1)d
d=30/m+1 _______(1)
given that T7/Tm-1= 5/9
a+(7-1+1)d/a+(m-1-1+1)d=5/9 (here , a+(7-1+1)d , +1 after 7-1 is because in the series the 1st A.M. is second in terms of series)
a+7d/a+(m-1)d=5/9
9a+63d = 5a+5md-5d
4a+68d =5md
4a= d(5m-68)
putting value of d frm (1)
by solving this you will get value of m as 14
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