Arithmetic Progressions: 10 Solved Problems for CBSE Class 10

Remember that feeling when you open your CBSE Class 10 Math paper and see a question from Arithmetic Progressions? Accha, sometimes it feels like a breeze, but sometimes, yaar, it can get tricky, right? Especially when you're trying to figure out if it's an $n^{th}$ term problem or a sum of $n$ terms one.

Don't worry, you're not alone! Arithmetic Progressions (AP) from NCERT Chapter 5 is super important, not just for your board exams but also for building a strong foundation for higher studies. It’s a chapter that often carries a good weightage, so mastering it is a smart move.

In this article, we're going to break down APs with 10 solved problems, just like your cool senior would explain. We'll cover everything from finding the $n^{th}$ term to calculating sums, and even tackle some real-life applications. So, suno, let's dive in and make AP your strong suit!

Before we jump into problem-solving, let's quickly recap the two main formulas that are your best friends in Arithmetic Progressions. These are the backbone of almost every problem you'll encounter in your CBSE Class 10 exams, whether from NCERT, RD Sharma, or RS Aggarwal.

**1. The $n^{th}$ term of an AP:** This formula helps you find any term in a sequence without listing all of them. Imagine you need the $50^{th}$ term of an AP, you can't just keep adding, right? This is where this formula comes in handy:

**2. Sum of the first $n$ terms of an AP:** What if you need to add up all the numbers in a long AP? This formula saves you a ton of time and effort:

Problem: Find the $10^{th}$ term of the AP: $2, 7, 12, ...$

Solution:
Given AP is $2, 7, 12, ...$
Here, the first term $a = 2$.

The common difference $d = 7 - 2 = 5$. (Also, $12 - 7 = 5$, so it's consistent.)
We need to find the $10^{th}$ term, so $n = 10$.

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

Substitute the values:
$a_{10} = 2 + (10-1)5$
$a_{10} = 2 + (9)5$
$a_{10} = 2 + 45$
$a_{10} = 47$

So, the $10^{th}$ term of the AP is $47$.

Problem: Find the sum of the first 20 terms of the AP: $1, 4, 7, 10, ...$

Solution:
Given AP is $1, 4, 7, 10, ...$
Here, the first term $a = 1$.

The common difference $d = 4 - 1 = 3$.
We need to find the sum of the first 20 terms, so $n = 20$.

Using the formula for the sum of the first $n$ terms: $S_n = \frac{n}{2}[2a + (n-1)d]$

Substitute the values:
$S_{20} = \frac{20}{2}[2(1) + (20-1)3]$
$S_{20} = 10[2 + (19)3]$
$S_{20} = 10[2 + 57]$
$S_{20} = 10[59]$
$S_{20} = 590$

Therefore, the sum of the first 20 terms of the AP is $590$.

Problem: Which term of the AP $21, 18, 15, ...$ is $-81$? Also, is $0$ a term of this AP?

Solution:
Given AP is $21, 18, 15, ...$
First term $a = 21$.
Common difference $d = 18 - 21 = -3$.

**Part 1: Finding which term is $-81$.**
Let $a_n = -81$. We need to find $n$.
Using the formula: $a_n = a + (n-1)d$
$-81 = 21 + (n-1)(-3)$
$-81 - 21 = -3(n-1)$
$-102 = -3(n-1)$
$\frac{-102}{-3} = n-1$
$34 = n-1$
$n = 34 + 1$
$n = 35$
So, the $35^{th}$ term of the AP is $-81$.

**Part 2: Is $0$ a term of this AP?**
Let $a_n = 0$. We need to find $n$.
$0 = 21 + (n-1)(-3)$
$-21 = -3(n-1)$
$\frac{-21}{-3} = n-1$
$7 = n-1$
$n = 7 + 1$
$n = 8$
Since $n$ is a positive integer, $0$ is the $8^{th}$ term of this AP.

This problem is a classic for CBSE Class 10 board exams, often appearing for 2-3 marks.

Problem: A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are $2 \frac{1}{2}$ m apart, how many rungs are there? What is the total length of wood required for the rungs?

Solution:
This is a classic real-life application of AP! Let's break it down.

Part 1: How many rungs are there?
Distance between top and bottom rungs $= 2 \frac{1}{2}$ m $= 2.5 \times 100$ cm $= 250$ cm.
Distance between consecutive rungs $= 25$ cm.

Number of gaps between rungs $= \frac{\text{Total distance}}{\text{Distance between rungs}} = \frac{250}{25} = 10$.

If there are 10 gaps, there must be $10 + 1 = 11$ rungs. (Think of 2 rungs, 1 gap; 3 rungs, 2 gaps, etc.)
So, $n = 11$.

Part 2: Total length of wood required for the rungs.
The lengths of the rungs form an AP.
First term $a$ (length of the bottom rung) $= 45$ cm.
Last term $l$ (length of the top rung) $= 25$ cm.
Number of rungs $n = 11$.

We need to find the sum of the lengths of all rungs, $S_n$.
Using the formula $S_n = \frac{n}{2}[a + l]$
$S_{11} = \frac{11}{2}[45 + 25]$
$S_{11} = \frac{11}{2}[70]$
$S_{11} = 11 \times 35$
$S_{11} = 385$ cm

Total length of wood required for the rungs is $385$ cm, or $3.85$ meters.

You've made it this far, which means you're serious about acing your math exams! Here's a quick recap of what we've covered:

* AP Basics: An Arithmetic Progression is a sequence where the difference between consecutive terms is constant, called the common difference ($d$).
*The $n^{th}$ Term:** Use $a_n = a + (n-1)d$ to find any term in the sequence.
*Sum of $n$ Terms:** Use $S_n = \frac{n}{2}[2a + (n-1)d]$ or $S_n = \frac{n}{2}[a + l]$ to find the sum of terms.
* Practice is Key: Consistent practice from NCERT and supplementary books like RD Sharma is crucial.
* Mindset Matters: Don't fear mistakes; embrace them as learning opportunities. Keep practicing, and you'll see improvement!