Question

What is the 24th term of the arithmetic sequence where a1 = 8 and a9 = 56?

Answer 1

Explanation:

Here we have the first term but must find the common ratio, r.

The nth term is a(n) = 8(r)^(n - 1).

To find r, substitute 9 for n in the above, obtaining 8(r)^(9-1), or

a(9) = 8(r)^8 = 56.

Solving for r^8, we get 56/8, or just 7.

Then r^8 = 7, and so r = 7^(1/8).

Thus, a(n) = 8(7)^(1/8)^(n-1), and so

a(24) = 8(7)^(23/8), or 2151.53

Answer 2

the formula is Tn = a + (n-1)d for arithmetic sequence

where Tn is the term, a is the first term, n is the no. of term, d is the difference

a = 8

Using a9=56,

56 = 8 + (9-1)d

48 = 8d

d = 6

thus, Tn = 8 + (n-1)6

when n=24, Tn = 8 + (24-1)(6) = 146