What is the 32nd term of the arithmetic sequence where a1 = 4 and a6 = 29 ?

Answer 1

An arithmetic sequence is one where the difference between a term and the next one is a constant. I like to think of it as a step size: the distance from a₁ to a₂ or a₅ to a₆ is the same.

So to go from a₁ to a₆ we've taken 5 steps, and covered a distance of a₆-a₁=29-4 = 25.

As the steps are all the same size, our step size is 25/5 = 5.

So to find a₃₂ we could just keep adding 5 over and over again, but note that this is 31 steps beyond a₁. So

a₃₂= a₁+ 5*(31 steps) = 4 + 155 = 159.