Question

What is the 32nd term of the arithmetic sequence where a1 = 4 and a6 = 24 ?

Answer 1

Use the explicit definition to find the common difference d\[a _{n}=a _{1}+(n-1)d\]\[24=4+(6-1)d\]\[d=4\]Then find the 32nd term\[a _{32}=4+(32-1)*4=128\]

Answer 2

So you could solve this without really using a formula because if a1 is 4 and a6 is 24 then d=4 because 6*4=24, so 6*32=192, so a32=192

Answer 3

yep