The third term of an A P is 6 and the fifth ter... - JAMB Mathematics 2023 Question

The nth term \(a_n\) of an arithmetic progression (A.P.) is given by the formula:

\[a_n = a + (n-1)d\]

where:
- \(a\) is the first term,
- \(n\) is the term number,
- \(d\) is the common difference.

Given that the third term \(a_3 = 6\) and the fifth term \(a_5 = 12\), we can set up a system of equations to solve for the first term \(a\) and the common difference \(d\):

1. For the third term:
\[a_3 = a + 2d = 6\]

2. For the fifth term:
\[a_5 = a + 4d = 12\]

Solving this system will give us the values of \(a\) and \(d\). Once we have those, we can use the formula for the sum of the first \(n\) terms of an A.P.:

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

to find the sum of the first twelve terms.

Let's solve for \(a\) and \(d\):

From equation (1):
\[a + 2d = 6\]

From equation (2):
\[a + 4d = 12\]

Subtracting equation (1) from equation (2):
\[(a + 4d) - (a + 2d) = 12 - 6\]
\[2d = 6\]

Dividing both sides by 2:
\[d = 3\]

Now substitute \(d = 3\) back into equation (1):
\[a + 2(3) = 6\]
\[a + 6 = 6\]
\[a = 0\]

So, \(a = 0\) and \(d = 3\).

Now, we can find the sum of the first twelve terms using the formula:
\[S_{12} = \frac{12}{2}[2(0) + (12-1)(3)]\]

Calculating this expression will give us the sum.

\[S_{12} = \frac{12}{2}[2(0) + 11(3)] = 6 \times 33 = 198\]