Question

What is the 41st term of the arithmetic sequence where a1 = 18 and a15 = –38 ?

Answer 1

18+14x = -38
x= -4
so:
a41 = 18 -4(40)

Answer 2

From the diagram, we see that if you want to get the 15th term from the 1st term, you need to add 'd' for 14 times
Therefore,
a1 + 14d = a15
18 + 14d = -38
d= (-38-18)/14 = -4

To get the 41st term, you can use the general formula
T(n) = a +(n-1)d
we get a =18, n =41, and d=-4
So T(41) = 18 +(41-1)(-4) = 18 - 160 = can you work out the answer?