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