absolutely my pleasure. i still love hearing from all of you all after all these years. it makes me happy. i'll never get the chance to meet most any of you but we are still friends.... just from a distance.
It would indeed- Sn = n/2[2a + (n − 1) × d] would give you 290=10/2[2a + (9) x 2] and solving for a (the first term) would give a=20, which is what Patrick got. This is also essentially the short hand formula for what Patrick derived at 6:14 onwards and can be used if you have e.g. 1000 pieces of wood.
Patrick, thanks for all you do.
absolutely my pleasure. i still love hearing from all of you all after all these years. it makes me happy. i'll never get the chance to meet most any of you but we are still friends.... just from a distance.
Would arithmetic progression formula work
It would indeed- Sn = n/2[2a + (n − 1) × d] would give you 290=10/2[2a + (9) x 2] and solving for a (the first term) would give a=20, which is what Patrick got.
This is also essentially the short hand formula for what Patrick derived at 6:14 onwards and can be used if you have e.g. 1000 pieces of wood.
thanks for explaining this. i was also going to show this formula but i wasn't sure if it would have been too much and what was best.
who did Patrick's homework =))...
patrickJMT itself.
What is this IB exam?
international baccalaureate exam