Hey guys and gals,

 

Let me put the qustion ' Tricky No. 1' here itself :

 

There are 6t terms in an A.P.

 

 Sum of first 2t terms : S1

 Sum of first 4t terms : S2

 Sum of last 2t terms : S3

 

If : (S1 * S2 * S3) - (S1^2 * S3) = 4

 

then (S2 - S1) is not less than :

 

a) 2^1/3

b) 2^2/3

c) 2^3/2

d) 2^5/2                 Pleazz explain :

 

 

 

Body A thrown upwards with a velocity 'u' making 1 with the ground collides with another body B, which is thrown with a velocity ' v ', from the other side, making an angle 2 with the ground. If they meet after time 't', at a horizontal distance of 'a' and 'b' along x-axis from bodies A and B respectively, then :

                           t = x / ucos1 = y / vcos2

 

t1 is the time of flight when body is projected at  with the horixontal and t2 is the time of flight when the body is projected at (90 - ) then :

 

=>  t1 * t2 = 2R / g

=>  t1 / t2 = tan

=>  The heights H1 and H2 during t1 and t2 relates R as : R = 4(H1 * H2)    

=>  H1 / H2 = tan²

 

body uniformily accelerares from rest at a rate a1 for time t1 and then at a2 for t2, at a3 for t3 .........

 

Then the average acceleration is : (a1t1 + a2t2 + a3t3 + ....) / (t1 + t2 + t3 +....)

 

 

for a vertical projectile : y = usin - gt² / 2

                                          = xtan - gx² / 2u²cos²

                                       H = t²(g / 8)

 

body dropped from the top of a building of height 'h'. Another body is thrown upwards from the ground level with a velocity 'u' making  with the ground.

                                                 OR

body thrown horixontally downwards with a velocity ' v ' from a height 'h' while another body is thrown upwards from the ground level with a velocity 'u' making  with the ground.

 

 

                                  

body accelerates at a constant rate 'a' for sometime from rest, and then with a constant velocity ' v ' moves for sometime, and retards at a rate 'r' coming to rest covering a distance ' x ',

                   then the total time taken = (x/v) + v/2(1/a + 1/r)

 

body is dropped from the top of a tower and after 'n' seconds another stone is thrown downwards with a velocity ' v '. They can meet after a time of :

                                   { (gn/2 - v) / (gn - v) }n

 

A bus accelerates at a constant rate 'a' from rest. A man standing at a distance of 'd' away from the bus should run at a minimum velocity of                v = (2ad) to get to the bus.   

 

 

 

 

 body dropped from height ' h ' so that it reaches the ground in time ' t ' such that final velocity is ' u '.         n  R

 

At height of (h - h/n) from the ground : the velocity is u / n

                                                     : the time taken is t / n   

 

body dropped from height ' h ' while at the sametime another body vertically

projected upwards with a velocity ' u '. these meet after a time of (h / u) at a

distance of (g / 2)(h / u)²  from the top.

 

' n ' balls are thrown every second such that, one ball thrown when the previous one reaches its maximum height. Now this max. height = (g / 2n² )

 

body dropped from a height and the time taken to cover successive 1 m distances are in the ratio is 1 : (2 - 1) : (3 - 2) : (4 - 3) : (5 - 4) : .......

 

body starts from rest and constantly accelerates at 'a' m/s² for time t1 and then retards at 'r' m/s² for time t2, coming to rest. Let t1 + t2 = T

Maximum velocity attained = at1 = rt2 = ( arT ) / (a + r) 

Total diatance covered = (arT² ) / 2(a + r) 

 

 

An arithmetic progression consisting up of ' 6t ' positive real terms is given.

 

 

If                                       (S1.S2.S3) - (S1².S3) = 4

 

 

d) 2^5/2     You need to explain your answer :

 

                                         Thathwamasi

 

 

 

 

ANS : I believe you all know this:

 

 

 

If there is any mistake, please let me know

If right, rate me kya??

 

                                              Thathwamasi

Shubham is absolutely right.

 

 

 

 

                                              Thathwamasi