Finding an Arithmetic Progression Given Two Terms

An Arithmetic Progression (A.P.) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by \(d\). The general form of an A.P. is \(a, a+d, a+2d, a+3d, \dots\), where \(a\) is the first term and \(d\) is the common difference.

The \(n\)th term of an A.P. is given by the formula:

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

We are given information about two terms of the A.P.:

• The 3rd term is 18. This means \(a_3 = 18\).
• The 12th term is 72. This means \(a_{12} = 72\).

Using the formula for the \(n\)th term, we can write these two pieces of information as equations:

For the 3rd term (\(n=3\)):

\(a_3 = a + (3-1)d\)

\(18 = a + 2d\) (Equation 1)

For the 12th term (\(n=12\)):

\(a_{12} = a + (12-1)d\)

\(72 = a + 11d\) (Equation 2)

Now we have a system of two linear equations with two variables, \(a\) and \(d\):

1. \(a + 2d = 18\)
2. \(a + 11d = 72\)

We can solve this system to find the values of \(a\) and \(d\). A common method is to subtract Equation 1 from Equation 2:

\((a + 11d) - (a + 2d) = 72 - 18\)

\(a + 11d - a - 2d = 54\)

\(9d = 54\)

Divide both sides by 9 to find \(d\):

\(d = \frac{54}{9}\)

\(d = 6\)

So, the common difference is 6. Now substitute the value of \(d\) back into Equation 1 to find the first term \(a\):

\(a + 2d = 18\)

\(a + 2(6) = 18\)

\(a + 12 = 18\)

Subtract 12 from both sides to find \(a\):

\(a = 18 - 12\)

\(a = 6\)

The first term \(a\) is 6.

Now that we have the first term (\(a=6\)) and the common difference (\(d=6\)), we can write out the terms of the A.P.:

• 1st term (\(a_1\)) = \(a = 6\)
• 2nd term (\(a_2\)) = \(a + d = 6 + 6 = 12\)
• 3rd term (\(a_3\)) = \(a + 2d = 6 + 2(6) = 6 + 12 = 18\) (Matches the given information)
• 4th term (\(a_4\)) = \(a + 3d = 6 + 3(6) = 6 + 18 = 24\)
• 12th term (\(a_{12}\)) = \(a + 11d = 6 + 11(6) = 6 + 66 = 72\) (Matches the given information)

The Arithmetic Progression is 6, 12, 18, 24, ...

Let's compare this A.P. with the given options:

Option (a): 6, 12, 18 ....

Option (b): 0, 9, 18 ....

Option (c): 12, 15, 18 ....

Option (d): 4.5, 9, 18 ....

The A.P. we found is 6, 12, 18, ..., which matches Option (a).

Revision Table: Understanding A.P. Concepts

Additional Information on Arithmetic Progressions

Arithmetic Progressions are a fundamental concept in sequences and series. They have many real-world applications, such as in simple interest calculations, calculating distances covered with constant acceleration (in discrete time steps), and even in patterns found in nature like the arrangement of leaves on a stem (phyllotaxis, though this can also involve Fibonacci sequences, which relate to A.P. in certain ways).

To find an A.P., you typically need to determine the first term (\(a\)) and the common difference (\(d\)). Once you have these two values, the entire sequence is defined. If you are given any two terms of the A.P., you can always set up a system of two linear equations using the \(n\)th term formula, as demonstrated in this problem, and solve for \(a\) and \(d\).

The sum of the first \(n\) terms of an A.P. is given by the formula:

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

or

\(S_n = \frac{n}{2} (a_1 + a_n)\)

where \(a_1\) is the first term and \(a_n\) is the \(n\)th term.