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?

Question

2191 views around the world

You can reuse this answer
Creative Commons License

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.!