The fifth term of an arithmetic sequence is 23 and the 12th term is 72. Determine the first three terms of the sequence and the nth term

Answer 1

Let's denote the first term of the arithmetic sequence as 'a' and the common difference as 'd'.
The fifth term of the sequence can be expressed as:
a + 4d = 23 (equation 1)
The twelfth term of the sequence can be expressed as:
a + 11d = 72 (equation 2)

Now, let's solve these two equations to find the values of 'a' and 'd'.

Multiplying equation 1 by 11 and equation 2 by 4, we get:

11(a + 4d) = 11(23) -> 11a +44d = 253 (equation 3)
4(a + 11d) = 4(72) -> 4a + 44d = 288 (equation 4)

Subtracting equation 3 from equation 4 to eliminate 'd', we get:

4a + 44d - 11a - 44d = 288 - 253
-7a = 35
a = -5

Now, substitute the value of 'a' in equation 1 to solve for 'd':

-5 + 4d = 23
4d = 28
d = 7

Therefore, the first term of the sequence (a) is -5 and the common difference (d) is 7.

The first three terms of the sequence are:
Term 1: a = -5
Term 2: a + d = -5 + 7 = 2
Term 3: a + 2d = -5 + 2(7) = 9

The nth term of the sequence can be represented as:
a + (n-1)d

So, the nth term is -5 + (n-1)(7) = -5 + 7n - 7 = 7n - 12.