A.P. Math Problems: Finding The Sum Of Terms

Hey there, math enthusiasts! Today, we're diving into the fascinating world of Arithmetic Progressions (A.P.s). These sequences, where the difference between consecutive terms is constant, pop up in all sorts of places, from financial calculations to scientific models. We'll be tackling a couple of common problems involving A.P.s, specifically focusing on how to find the sum of a certain number of terms when you're given information about specific terms in the sequence. So, grab your calculators and let's get started!

Understanding Arithmetic Progressions

Before we jump into solving, let's quickly recap what an A.P. is. An Arithmetic Progression is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference, usually denoted by 'd'. The first term of an A.P. is denoted by 'a'.

The formula for the n-th term of an A.P. is given by:

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

where $a_n$ is the n-th term, $a$ is the first term, and $d$ is the common difference.

Now, when it comes to finding the sum of the first 'n' terms of an A.P., we have two handy formulas:

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

2. $$S_n = \frac{n}{2} (a + a_n)$$

These formulas are incredibly useful. The first one uses the first term and the common difference, while the second one uses the first term and the last term ($a_n$) of the sum. We'll be using these to solve our problems.

Problem (a): Finding the Sum of the First 30 Terms

Let's tackle our first problem: In an A.P., the 10th term is 29 and the 20th term is 59. Find the sum of the first 30 terms?

Here, we are given two specific terms of the A.P. and asked to find the sum of the first 30 terms. This means we need to find the first term ('a') and the common difference ('d') first. We can use the formula for the n-th term to set up a system of equations.

We are given:

• $a_{10} = 29$
• $a_{20} = 59$

Using the formula $a_n = a + (n-1)d$, we can write:

For the 10th term:

$$a_{10} = a + (10-1)d = a + 9d = 29 ext{ (Equation 1)}$$

For the 20th term:

$$a_{20} = a + (20-1)d = a + 19d = 59 ext{ (Equation 2)}$$

Now, we have a system of two linear equations with two variables, 'a' and 'd'. We can solve this system using elimination or substitution. Let's use elimination by subtracting Equation 1 from Equation 2:

$$(a + 19d) - (a + 9d) = 59 - 29$$

$$a + 19d - a - 9d = 30$$

$$10d = 30$$

$$d = \frac{30}{10} = 3$$

So, the common difference (d) is 3.

Now that we have 'd', we can substitute it back into either Equation 1 or Equation 2 to find 'a'. Let's use Equation 1:

$$a + 9d = 29$$

$$a + 9(3) = 29$$

$$a + 27 = 29$$

$$a = 29 - 27 = 2$$

Thus, the first term (a) is 2.

We have successfully found the first term and the common difference: $a = 2$ and $d = 3$. Now we can find the sum of the first 30 terms ($S_{30}$) using the formula $S_n = \frac{n}{2} [2a + (n-1)d]$.

$$S_{30} = \frac{30}{2} [2(2) + (30-1)(3)]$$

$$S_{30} = 15 [4 + (29)(3)]$$

$$S_{30} = 15 [4 + 87]$$

$$S_{30} = 15 [91]$$

To calculate $15 \times 91$: $15 \times 90 = 1350$ $15 \times 1 = 15$ $1350 + 15 = 1365$

Therefore, the sum of the first 30 terms of this A.P. is 1365.

This problem highlights the power of using the n-th term formula to unlock the fundamental properties (first term and common difference) of an A.P., which then allows us to calculate any sum we desire. It's like deciphering a code where each piece of information helps you understand the whole sequence.

Problem (b): Another A.P. Summation Challenge

Let's move on to our second problem, which is quite similar in structure: In an A.P., the 12th term is 27 and the 25th term is 53. Find the sum of the first 40 terms?

Again, we're given two terms and asked for a sum. The strategy remains the same: find 'a' and 'd' first.

We are given:

• $a_{12} = 27$
• $a_{25} = 53$

Using the n-th term formula $a_n = a + (n-1)d$:

For the 12th term:

$$a_{12} = a + (12-1)d = a + 11d = 27 ext{ (Equation 3)}$$

For the 25th term:

$$a_{25} = a + (25-1)d = a + 24d = 53 ext{ (Equation 4)}$$

Let's solve this system of equations. Subtract Equation 3 from Equation 4:

$$(a + 24d) - (a + 11d) = 53 - 27$$

$$a + 24d - a - 11d = 26$$

$$13d = 26$$

$$d = \frac{26}{13} = 2$$

So, the common difference (d) is 2.

Now, substitute $d = 2$ into Equation 3 to find 'a':

$$a + 11d = 27$$

$$a + 11(2) = 27$$

$$a + 22 = 27$$

$$a = 27 - 22 = 5$$

Therefore, the first term (a) is 5.

We have $a = 5$ and $d = 2$. Now we need to find the sum of the first 40 terms ($S_{40}$) using the formula $S_n = \frac{n}{2} [2a + (n-1)d]$.

$$S_{40} = \frac{40}{2} [2(5) + (40-1)(2)]$$

$$S_{40} = 20 [10 + (39)(2)]$$

$$S_{40} = 20 [10 + 78]$$

$$S_{40} = 20 [88]$$

To calculate $20 \times 88$: $20 \times 80 = 1600$ $20 \times 8 = 160$ $1600 + 160 = 1760$

Thus, the sum of the first 40 terms of this A.P. is 1760.

These types of problems are excellent for building your confidence with algebraic manipulation and understanding the foundational principles of sequences. By breaking down the problem into finding the common difference and the first term, you can systematically solve for any sum or term within an arithmetic progression. It's a core skill in many areas of mathematics and beyond.

Conclusion: Mastering A.P. Sums

We've successfully navigated two problems involving arithmetic progressions, demonstrating how to find the sum of terms when given specific term values. The key takeaway is to always first determine the first term (a) and the common difference (d) using the formula for the n-th term. Once you have these values, calculating the sum of any number of terms becomes a straightforward application of the summation formulas.

Arithmetic progressions are fundamental building blocks in mathematics. They're not just abstract concepts; they have practical applications in finance (like calculating loan payments or annuities), physics (like motion with constant acceleration), and computer science (like analyzing algorithms). Mastering these problems means you're building a solid foundation for understanding more complex mathematical ideas.

If you'd like to explore more about arithmetic progressions and related concepts, I highly recommend checking out the resources at Khan Academy. They offer a wealth of free educational materials, including detailed explanations and practice exercises on sequences and series. For a more in-depth theoretical understanding, the Wikipedia page on Arithmetic Progression provides a comprehensive overview.