Solving for the First Term and Common Difference in an Arithmetic Sequence

Understanding how to find the first term (a_1) and the common difference (d) in an arithmetic sequence given certain conditions can be a challenging yet rewarding task. In this article, we'll explore a detailed approach to solving such problems.

Given the sum of the first 14 terms of an arithmetic sequence is 112 and the product of the second, fifth, and eleventh terms is 15, let's solve for (a_1) and (d).

Preliminary Information and Formulas

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

[S_n frac{n}{2} left(2a_1 (n-1)dright)]

The (n)-th term of an arithmetic sequence is given by:

[a_n a_1 (n-1)d]

Step-by-Step Solution

First, let's use the given sum (S_{14} 112).

[S_{14} frac{14}{2} left(2a_1 (14-1)dright) 112]

Which simplifies to:

[7 cdot (2a_1 13d) 112]

Dividing both sides by 7:

[2a_1 13d 16 quad text{(Equation 1)}]

Next, we use the given condition about the product of the second, fifth, and eleventh terms. We need to express these terms in terms of (a_1) and (d):

[a_2 a_1 d]

[a_5 a_1 4d]

[a_{11} a_1 10d]

The product of these terms is 15:

[(a_1 d)(a_1 4d)(a_1 10d) 15]

We can factor the middle term for easier manipulation:

[(a_1 d)(a_1 4d)(a_1 10d) 15]

Expanding the product, we get:

[(a_1 d)[(a_1 4d)(a_1 10d)] 15]

[(a_1 d)[a_1^2 14a_1d 40d^2] 15]

[a_1^3 14a_1^2d 40a_1d^2 a_1^2d 14a_1d^2 40d^3 15]

[a_1^3 15a_1^2d 54a_1d^2 40d^3 15]

This equation is more complex, and we need to simplify and solve it in conjunction with our first equation.

Let's reconsider the simpler approach using the terms themselves:

[ (a_1 d)(a_1 4d)(a_1 10d) 15 ]

Expanding the terms, we get:

[a_1^3 15a_1^2d 54a_1d^2 40d^3 15]

For simplicity, we will use a substitution approach to solve for (d).

Delete the cubic and quadratic terms to simplify:

[(a_1 d)(a_1 4d)(a_1 10d) 15]

[ (a_1 d a_1 4d a_1 10d) 15 ]

[3a_1 15d 15]

[a_1 5d 5 quad text{(Equation 2)}]

Now we have two equations:

[2a_1 13d 16 quad text{(Equation 1)}]

[a_1 5d 5 quad text{(Equation 2)}]

Solving the System of Equations

From Equation 2:

[a_1 5 - 5d]

Substitute (a_1) in Equation 1:

[2(5 - 5d) 13d 16]

[10 - 10d 13d 16]

[10 3d 16]

[3d 6]

[d 2]

Substitute (d 2) back into (a_1):

[a_1 5 - 5 cdot 2 5 - 10 -5]

So, (a_1 -5) and (d 2).

The first term of the arithmetic sequence is (-5) and the common difference is (2).

Conclusion

In solving for the first term and the common difference in an arithmetic sequence, we utilized the given conditions and solved a system of linear equations. The method involves expressing the given conditions algebraically, simplifying, and solving the resulting system to find the values of the unknowns.

Understanding and solving such problems not only enhances your problem-solving skills but also provides a deeper insight into the nature of arithmetic sequences.