Question

Solve this..sum...........

Answer 1

if p q r are in GP then
q^(2) = p x r
now given equation is
p(x^2) + 2qx +r

by discriminate formula
$x \: = \frac{ - b \: \: \: \: ( + )or( - ) \sqrt{ {b}^{2} - 4ac } }{2a}$
we get
x = [-2q (+or-)√{4(q^2) - 4(p)(r)}/2p
filling q^2 = pr
x= -2q/2p
x = -q/p
now as both the equations have same roots
x will satisfy both equations
so
d(-q/p)^2 + 2e(-q/p) + f = 0
now
taking second term on R.H.S
so
d(q^2) / (p^2) +f = 2eq/p
multiply whole equation by p^2
d(q^2) + f(p^2) = 2eqp
divide whole equation by q
dq + f(p^2)/q = 2ep
multiply whole equation by r
dqr + f(p)(p)(r)/q = 2epr
as pr = q^2
dqr + fpq = 2epr
divide whole the equation by pqr
d/p + f/r = 2e/q
which shows
d/p , e/q , f/r are in A.P

Answer 2

hey there,

Since p, q, r are in a A.P.
we ave (q -p ) = ( r - q ) = common difference

Now consider any G.P with first term = A, and common ratio = R
then
the pth , qth and rth trms of that G.P will be
A * R^(p-1), A * R^(q-1), A * R^(r-1)

Find the ratio of the first two terms above

Ratio = A R^(q-1) / AR^(p-1) = = R^(q-p)

Also , the other ratio of the second two terms
Ratio = A R^(r-1) / AR^(q-1) = = R^(r-q)

Both the above ratios are equal since ( q - p ) = ( r - p )
Hence the said terms are in a G.P


Hope this helps!