The sum of the first 3 terms of a sequence is 9. The sum of the first 6 terms is -63. Given that this is an arithmetic sequence, find the first term and the common difference.

1. Problem Description

The sum of the first 3 terms of a sequence is

9. The sum of the first 6 terms is -

6

3. Given that this is an arithmetic sequence, find the first term and the common difference.

2. Solution Steps

Let $a$ be the first term and $d$ be the common difference of the arithmetic sequence.

The sum of the first $n$ terms of an arithmetic sequence is given by:

$S_n = \frac{n}{2}[2a + (n-1)d]$

Given that the sum of the first 3 terms is 9:

$S_3 = \frac{3}{2}[2a + (3-1)d] = 9$

$\frac{3}{2}(2a + 2d) = 9$

$3(a+d) = 9$

$a+d = 3$ (1)

Given that the sum of the first 6 terms is -63:

$S_6 = \frac{6}{2}[2a + (6-1)d] = -63$

$3(2a + 5d) = -63$

$2a + 5d = -21$ (2)

Now we have a system of two linear equations with two variables:

$a+d = 3$ (1)

$2a + 5d = -21$ (2)

From equation (1), we can express $a$ in terms of $d$:

$a = 3 - d$

Substitute this into equation (2):

$2(3-d) + 5d = -21$

$6 - 2d + 5d = -21$

$3d = -27$

$d = -9$

Now, substitute $d=-9$ back into equation (1):

$a + (-9) = 3$

$a = 3 + 9$

$a = 12$

3. Final Answer

The first term is 12 and the common difference is -

9. $a = 12, d = -9$