The problem given is:

If the sum of n terms of an AP is $3n^2 - n$, and its common difference is 6, then its first term is?

Let's solve this step by step.

Step 1: Write the Sum Formula for an AP

The sum of the first $n$ terms of an arithmetic progression (AP) can be written as: $S_n = 3n^2 - n$

The general formula for the sum of the first $n$ terms of an AP is: $S_n = \frac{n}{2} \times [2a + (n-1)d]$ where:

• $a$ is the first term,
• $d$ is the common difference,
• $S_n$ is the sum of the first $n$ terms.

Step 2: Compare and Find Relations

We are given the common difference $d = 6$.

Now, we compare the given sum formula $S_n = 3n^2 - n$ with the general formula.

Step 3: Expand and Equate

Using the general formula:

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

Substituting $d = 6$:

$S_n = \frac{n}{2} [2a + (n-1) \times 6]$

$S_n = \frac{n}{2} [2a + 6n - 6]$

Expanding:

$S_n = \frac{n}{2} \times (2a + 6n - 6)$

$S_n = n(a + 3n - 3)$

$S_n = 3n^2 + n(a - 3)$

Step 4: Compare with Given Sum

We have:

$S_n = 3n^2 - n$

Now, compare:

$3n^2 + n(a - 3) = 3n^2 - n$

Step 5: Solve for $a$

To solve for $a$, compare the coefficients of like terms of $n$:

For the linear term ($n$): $a - 3 = -1$

$a = 2$

Final Answer

The first term of the AP is $a = 2$.

Would you like me to provide more details or have any questions? Here are some related questions to explore:

1. How can we find the $n$-th term of an AP if we know the first term and common difference?
2. How does changing the common difference affect the sum formula for an AP?
3. What is the relationship between the sum formula and the general term of an AP?
4. How do we determine the sum of an infinite arithmetic progression?
5. How does this problem change if the sum formula was different?

Tip: Always compare coefficients when equating polynomial expressions for an unknown value.