 Hello and welcome to the session. Let's work out the following problem. It says a particle moving in a straight line with uniform acceleration describes successive equal distances in times T1, T2 and T3 prove that 1 upon T1 minus 1 upon T2 plus 1 upon T3 is equal to 3 upon T1 plus T2 plus T3. So, let's now move on to the solution. We are given that a particle moving in a straight line describes the equal distances in times T1, T2 and T3. So, let A B is equal to, B C is equal to, C D is equal to S be the three equal distances and let U be the initial velocity A meter per second square is the uniform acceleration. Now, we have A B is equal to, B C is equal to, C D is equal to S particle is moving in a straight line describing equal distances in times T1, T2 and T3. Now, the particle describes the distance A B is equal to S in time T1. Similarly, it describes distance A C that is this which is equal to S plus S that is 2 S in time T1 plus T2. Since we are given that the particle describes the equal distances in time T1, T2 and T3. So, if it describes the distance S in time T1, so it describes the distance 2 S in time T1 plus T2 and also the distance A D that is equal to 3 S in time T1 plus T2 plus T3 right. Now, the distance traveled S is given by the formula U T plus 1 by 2 F T square, S is the distance traveled, U is the initial velocity and F is the acceleration must remember this formula and write it while using this. Now, the distance A B is S is given by U T1, T is the time taken plus 1 by 2 F T1 square. Since the time taken to describe the distance S is T1, so here we have used T1 and similarly distance A C that is 2 S is equal to U into T1 plus T2. Since the time taken to describe the distance A C that is 2 S is T1 plus T2 and here we have 1 by 2 into F into T1 plus T2 whole square and A D that is 3 S is equal to U into T1 plus T2 plus T3. Since the time taken to describe the distance 3 S is T1 plus T2 plus T3 to 1 by 2 T1 plus T2 plus T3 whole square. Let us name this as 1, this as 2 and this as 3. Now subtract 1 from 2, 2 from 3 we have subtracting 1 from 2 we have 2 S minus S is equal to U into T1 plus T2 plus 1 by 2 F into T1 plus T2 whole square minus U T1 plus 1 by 2 F T1 square. So 2 S minus S is equal to U T1 plus U T2 plus 1 by 2 F into T1 square plus T2 square plus 2 into T1 T2 applying the formula A plus B whole square minus U T1 minus 1 by 2 F T1 square. So this is equal to U T1 gets cancelled with U T1 1 by 2 F T1 square will get cancelled with minus 1 by 2 F T1 square we are left with U T2 plus 1 by 2 F T2 square plus 1 by 2 gets cancelled with 2 here and we have F T1 into T2. So this is equal to U T2 plus 1 by 2 F into T2 square plus T1 T2. Now we subtract 2 from 3 so we have 3 S minus 2 S is equal to 3 is U into T1 plus T2 plus T3 that is U T1 plus U T2 plus U T3 plus 1 by 2 F into T1 plus T2 plus T3 whole square minus AC that is 2 S was equal to U into T1 plus T2 that is U T1 plus U T2 plus 1 by 2 F into T1 plus T2 whole square. Now this is equal to U T1 plus U T2 plus U T3 plus 1 by 2 F into T1 square plus T2 square plus T3 square plus 2 T1 T2 plus 2 T2 T3 plus 2 T3 T1 minus now since we have negative sign of outside the bracket when we will open the bracket sign will change so this becomes minus U T1 minus U T2 minus 1 by 2 F into T1 plus T2 whole square is T1 square plus T2 square plus 2 into T1 T2 so again we have U T1 plus U T2 plus U T3 plus 1 by 2 F T1 square so we have opened the bracket now U T1 gets cancelled with minus U T1 U T2 gets cancelled with minus U T2 1 by 2 F T1 square gets cancelled with minus 1 by 2 F T1 square and we are left with U T3 plus 1 by 2 F T3 square plus F T2 T3 plus F T3 T1 this can be written as U T2 plus 1 by 2 into F taking T2 common we have 1 by 2 F T2 into T2 plus 2 T1 let's name this as A and this can be written as U T3 plus 1 by 2 taking 1 by 2 F common we have T3 square plus 2 T2 T3 plus 2 T3 T1 this is again equal to U T3 plus 1 by 2 F taking T3 common we have T3 plus 2 T2 plus 2 T1 now we are also 3 S minus 2 S is S S is equal to this and let's name this as B now from A and B we have S is equal to from A we have S is equal to U T2 plus 1 by 2 F T2 into T2 plus 2 T1 and from B we have S is equal to U T3 plus 1 by 2 F T3 into T3 plus 2 T2 plus 2 T1 now here dividing both sides by T2 we have S by T2 is equal to U plus 1 by 2 F into T2 plus T1 and here dividing both sides by T3 we have S upon T3 is equal to U plus 1 by 2 F into T3 plus 2 T2 plus 2 T1 let's name this as 4 and this as 5 and also from 1 we have S is equal to U T1 1 square plus 1 by 2 F T1 square now dividing both sides by T1 we have S T1 is equal to U here we have U T1 so here we have U plus 1 by 2 F T1 let's name this as 6 now we have to prove that 1 upon T1 minus 1 upon T2 plus 1 upon T3 is equal to 3 upon T1 plus T2 plus T3 this is what we have to prove so now we find S upon T1 minus S upon T2 plus S upon T3 so we have S upon T1 is U plus 1 by 2 F T1 minus S upon T2 which is this so this is U plus 1 by 2 into T1 plus S upon T3 which is this so this is U plus 1 by 2 plus 2 T2 plus 2 T1 minus S upon T2 plus S upon T3 is equal to U plus 1 by 2 F T1 minus U minus 1 by 2 plus 2 T1 plus U plus 1 by 2 F into T3 plus 2 T2 plus 2 T1 U gets cancelled with U and we have U plus in all these terms taking 1 by 2 F common so we have 1 by 2 into F into T1 minus of T2 plus 2 T1 plus 3 plus 2 T2 plus 2 T1 this is equal to U plus 1 by 2 F into T1 minus T2 minus 2 T1 plus T3 plus 2 T2 plus 2 T1 so this is equal to U plus 1 by 2 into F minus 2 T1 gets cancelled with plus 2 T1 and we have T1 2 T2 minus T2 is plus T2 plus T3 so this is equal to 3S upon T1 plus T2 plus T3 because we had AD which is equal to 3S is equal to U into T1 plus T2 plus T3 plus 1 by 2 F into T1 plus T2 plus T3 square so this implies dividing both sides by T1 plus T2 plus T3 we have 3S upon T1 plus T2 plus T3 is equal to U plus 1 by 2 into F into T1 plus T2 plus T3 so here we have substituted this by 3S upon T1 plus T2 plus T3 so we have S upon T1 minus S upon T2 plus S upon T3 is equal to 3S upon T1 plus T2 plus T3 now taking as common from here and cancelling on both sides we have 1 upon T1 minus 1 upon T2 plus 1 upon T3 is equal to 3 upon T1 plus T2 plus T3 and this is what we have to prove so this completes the question and the session by for now take care have a good day.