If 64, 27, 36 are the Pth Qth and Rth terms of a GP, then P + 2Q is equal to
R
2R
3R
4R
C.
3R
Let a be the first term and r be the common ratio of a GP.
Pth, Qth and Rth terms of a GP are respectively arP - 1, arQ - 1 and arR - 1.
According to question,
arP - 1 = 64 ...(i)
arQ - 1 = 27 ...(ii)
arR - 1 = 36 ...(iii)
Dividing Eq. (i) by Eq. (ii), we get
rP - Q = ...(iv)
r3Q - 3R = ...(v)
Multiplying Eq. (iv) and Eq. (v), we get
Let p, q and r be the sides opposite to the angles P, Q and R, respectively in a PQR. If r2sin(P)sin(Q) = pq, then the triangle is
equilateral
acute angled but not equilateral
obtuse angled
right angled
Let p, q and r be the side4s opposite to the angles P, Q and R, respectively in a PQR. Then, 2pr equals
p2 + q2 + r2
p2 + r2 - q2
q2 + r2 - p2
p2 + q2 - r2
Let P (2,-3) , Q -2 1) be the vertices of the PQR. If the centroid of PQR lies on the line 2x + 3y = 1, then the locus of R is
2x + 3y = 9
2x - 3y = 7
3x + 2y = 5
3x - 2y = 5
Let P be the mid-point of a chord joining the vertex of the parabola y2 = 8x to another point on it. Then, the locus of P is
y2 = 2x
y2 = 4x
= 1
The line x = 2y intersects the ellipse at the points P and Q. The equation of the circle with PQ as diameter is
x2 + y2 = 1
x2 + y2 = 2
x2 + y2 =
The eccentric angle in the first quadrant of a point on the ellipse at a distance units from the centre of the ellipse is
The transverse axis of a hyperbola is along the x - axis and its length is 2a. The vertex of the hyperbola bisects the line segment joining the centre and the focus. The equation of the hyperbola is
6x2 - y2 = 3a2
x2 - 3y2 = 3a2
x2 - 6y2 = 3a2
3x2 - y2 = 3a2