A particle moves in a straight line with constant acceleration, starting from a point on the line. It moves through distances a ,b,c and d, in 0. n, 2n and 3n s respectively. Show that,
(i) d-a = 3(c-b) (ii) initial velocity = `(4b-3a-c)/(2n)` and (iii) acceleration = `(c+a-2b)/(n^(2)).`

A particle moves along a straight line with uniform acceleration. If it travels through distances a and b, respectively in the l - th and m-th seconds then prove that the distance travelled in the n-th second is s_(n) = (a(n-m)+b(l-n))/(l-m)

If a variable tangent to the curve x^2y=c^3 makes intercepts a , b on x a n d y-a x e s , respectively, then the value of a^2b is (a) 27c^3 (b) 4/(27)c^3 (c) (27)/4c^3 (d) 4/9c^3

A particle crosses points A, B, C in a straight line at a constant acceleration. The particle travels from A to B in time t_(1) and B to C in time t_(2) . If AB = a , BC = b then prove that the acceleration of the particle is 2(bt_(1)-at_(2))//t_(1)t_(2)(t_(1)+t_(2)) .

A particle starting from a point A and moving with a positive constant acceleration along a straight line reaches another point B is time T. Suppose that the initial velocity of the particle is u gt0 and P is the midpoint of the line AB. If the velocity of the particle at point P is v_(1) and if the velocity at time (T)/(2) is v_(2) , then -

The line joining the points (-2,1,-8)a n d(a ,b ,c) is parallel to the line whose direction ratios are 6,2,a n d3. Find the values of a ,b and c

If the 3rd, 4th. 5th and sixth term in the expansion of (x+alpha)^n are a,b,c,d respectively, then prove that ((b^2-ac)/(c^2-bd))=(5a)/(3c).

In triangle A B C , if sinAcosB=1/4 and 3t a n A=t a n B ,t h e ncot^2A is equal to (a)2 (b) 3 (c) 4 (d) 5.

In triangle A B C ,points D , Ea n dF are taken on the sides B C ,C Aa n dA B , respectively, such that (B D)/(D C)=(C E)/(E A)=(A F)/(F B)=ndot Prove that triangle D E F=(n^2-n+1)/((n+1)^2) triangle(A B C)dot

In a triangle A B C , the bisectors of angles Ba n dC lies along the lines x=y a n d y=0. If A is (1,2) , then the equation of line B C is

Consider two lines L_1a n dL_2 given by a_1x+b_1y+c_1=0a n da_2x+b_2y+c_2=0 respectively where c1 and c2 !=0, intersecting at point PdotA line L_3 is drawn through the origin meeting the lines L_1a n dL_2 at Aa n dB , respectively, such that PA=P B . Similarly, one more line L_4 is drawn through the origin meeting the lines L_1a n dL_2 at A_1a n dB_2, respectively, such that P A_1=P B_1dot Obtain the combined equation of lines L_3a n dL_4dot