If the pth ,qth and rth terms of an A.P. be in G P with the common ratio $k$, then the roots of the equation $(q-r)x^{2}+(r-p)x+(p-q)=0 $ are
- $1$ and $k$
- $2$ and $k$
- $1$ and $\dfrac{1}{k}$
-
None of these
Given p, q, r terms of an AP are in GP with ratio k, we have a+(p-1)d = A, a+(q-1)d = Ak, a+(r-1)d = Ak^2. This implies (q-p)d = A(k-1) and (r-q)d = Ak(k-1). Thus (r-q)/(q-p) = k. The equation is (q-r)x^2 + (r-p)x + (p-q) = 0. Since the sum of coefficients is 0, x=1 is a root. The product of roots is (p-q)/(q-r) = 1/k.
Let the pth, qth and rth terms of the arithmetic progression be a plus p minus 1 times d, a plus q minus 1 times d and a plus r minus 1 times d. Since these terms are in geometric progression with common ratio k, we can write the equations a plus p minus 1 times d equals A, a plus q minus 1 times d equals Ak and a plus r minus 1 times d equals A times k squared. By subtracting the first equation from the second, we find q minus p times d equals A times the quantity k minus 1. By subtracting the second equation from the third, we find r minus q times d equals A times k times the quantity k minus 1. Dividing these two differences eliminates A and d, leaving k equals r minus q divided by q minus p. We substitute x equals 1 into the given quadratic and find it evaluates to 0, making 1 a root. Using the product of roots formula for the quadratic, the other root is p minus q divided by q minus r. Substituting r minus q equals k times q minus p into the product formula gives the second root as 1 divided by k. Therefore, the roots are 1 and 1 divided by k.