Between the roots of the quadratic equation #3px^2 -10px +5q#=0(#p>0,frac{q}p<frac{5}3#),an odd number of A.M's are inserted and their sum exceeds their number by 10.Find the number of A.M's inserted?

Let, #alpha and beta# are the roots of the given qdr. eqn.,

where, #alpha={10p+sqrt(100p^2-60pq)}/(6p)#,

#=(10p)/(6p)+{2sqrt(25p^2-15pq)}/(6p)#,

#=5/3+1/3sqrt{(25p^2-15pq)/p^2}," so that, "#

#alpha=5/3-1/3sqrt(25-15q/p), and, #

#beta=5/3+1/3sqrt(25-15q/p)#.

Let, the odd no. of AMs inserted btwn. #alpha and beta" be "2m-1#.

Let these AMs be #A_1,A_2,...,A_(2m-1),# so that,

#alpha, A_1,A_2,..,A_(2m-1),beta............(star)#, are in some AP.

But, we are Given that,

#A_1+A_2+...+A_(2m-1)=(2m-1)+10=2m+9#.

Accordingly, in the usual notation of an AP, we have,

#(2m-1)/2[2A_1+{(2m-1)-1}d]=2m+9, or, #

#(2m-1){A_1+(m-1)d}=2m+9, i.e., #

#(2m-1){(A_1-d)+md}=2m+9#.

Knowing that, #A_1-d=alpha#, we find that,

#(2m-1)(alpha+md)=2m+9....................(star_1)#.

#(star)# reminds us that, #beta# is the #(2m+1)^(th)# term of AP.

#:. beta=alpha+{(2m+1)-1}d, or, beta-alpha=2md#.

#:. 2md=2/3sqrt(25-15q/p), or, md=1/3sqrt(25-15q/p)#.

#:. (star_1)rArr(2m-1){alpha+1/3sqrt(25-15q/p)}=2m+9, i.e.,#,

Sub.ing the value of #alpha#, we get,

#(2m-1){5/3-1/3sqrt(25-15q/p)+1/3sqrt(25-15q/p)}=2m+9#.

#:. 5(2m-1)=3(2m+9)," giving, "m=8#.

This means that #(2m-1)=15# AMs have been inserted.

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