if m arthematic means are inserted between 1 and 31 so that the ratio of 7th and (m-1)th means is 5:9 then the value of m is

Dear Saifulla,

 please find the attachment below and solve the equation you get by simplifying

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Approved

Let the arithmetic mean inserted be A₁ , A₂ , A₃,....A_m.

now , series is 1,A₁ , A₂ , A₃,....A_{m ,} 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 /T_m-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

Approved