Between the roots of the quadratic equation 3px^2-10px+5q=0(p>0,q/p<5/3),an odd number of A.M.S are insertred and their sum exceeds their number by 10.find the number of A.M.S inserted?

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Answer 1 · 2017-08-27T04:42:13

# 15.#

Let, #alpha, and, beta# be the roots of the given quadr. eqn. :

#3px^2-10px+5q=0, (p>0, q/p<5/3).#

#:. alpha+beta=-(-10p)/(3p),,# and, as #p>0, i.e., p!=0,# we have,

# alpha+beta=10/3......................................................................(1).#

The No. pf AMs inserted btwn. #alpha, and, beta# is odd, let

this no. be #2k-1, k in NN.#

If we denote these AMs by #A_1,A_2,...,A_(2k-1),# by what is given,

we conclude that, #alpha, A_1,A_2,..., A_(2k-1), beta# are in AP.

Observe that the above AP has #(2k-1)+2=2k+1.#

To find its sum. we use the following Formula :

Sum=1/2(No. of Terms){First Term+Last Term}.

#:. alpha+A_1+A_2+...+A_(2k-1)+beta=1/2(2k+1)(alpha+beta)...(2).#

But, it is given that, #A_1+A_2+...+A_(2k-1)=(2k-1)+10.#

Using this, in #(2),# together with #(1),# we get,

# 10/3+{(2k-1)+10}=1/2(2k+1)(10/3), i.e., #

# 10/3+2k+9=5/3(2k+1), or,# multiplying by #3,#

# 10+6k+27=10k+5 rArr k=8,"giving, "#

the desired No. of Ams. is #2k-1=15.#

Enjoy Maths.!