CBSE Class 10 Maths: Arithmetic Progressions Exercise 5.2

Understanding the n-th Term Formula of an AP

The core of Exercise 5.2 is understanding how an Arithmetic Progression grows. If the first term is $a$ and the common difference is $d$, the terms are written as $a_1 = a$, $a_2 = a + d$, $a_3 = a + 2d$, and $a_4 = a + 3d$. Notice the pattern: the coefficient of $d$ is always one less than the position of the term. Therefore, for the $n$-th term ($a_n$), the coefficient of $d$ is $(n-1)$. This gives us our fundamental formula:

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

Here, $a_n$ is the general term (or last term, sometimes denoted as $l$), $a$ is the first term, $n$ is the total number of terms (which must always be a positive integer), and $d$ is the common difference. If you are given any three of these four variables, you can easily find the fourth by setting up a simple linear equation.

Step-by-Step Blueprint to Solve Exercise 5.2 Problems

1. Identify and List the Givens — Read the problem carefully to extract the values of $a$ (first term), $d$ (common difference), $n$ (term position), or $a_n$ ($n$-th term). Make sure to check if the common difference is positive or negative.
2. Substitute into the Formula — Write down the general formula $a_n = a + (n-1)d$ on your sheet and substitute the known parameters. Using parentheses around negative numbers prevents sign mistakes.
3. Solve for the Unknown Variable — Isolate the unknown variable using algebraic operations. If solving for $d$ or $n$, simplify the expression inside the bracket before dividing or subtracting.
4. Verify the Nature of 'n' — If you are solving for $n$ (the number of terms), remember that $n$ must always be a positive integer ($1, 2, 3, \dots$). If you calculate a fraction or negative number for $n$, the given value cannot be a term of the AP.

Pro-Tips and Common Board Exam Mistakes

Many students lose marks in Exercise 5.2 due to simple algebraic errors. Here is how to avoid them:

1. The Negative Difference Trap: When the common difference $d$ is negative, always use parentheses when multiplying. Writing $a + (n-1)-3$ instead of $a + (n-1)(-3)$ often leads to accidental subtraction instead of multiplication.

1. Finding the $n$-th Term from the End: To find the $n$-th term from the end of a finite AP, you can reverse the AP (making the last term $l$ your new first term and $-d$ your new common difference) or use the shortcut formula:

$\text{Term from the end} = l - (n-1)d$

Practice Questions with Solutions

• Q: Find the 11th term of the AP: $-3, -1/2, 2, \dots$ A: Step 1: Identify the given values from the AP. First term $a = -3$. Common difference $d = -1/2 - (-3) = -1/2 + 3 = 5/2$. We need to find the 11th term, so $n = 11$. Step 2: Use the formula $a_n = a + (n-1)d$. $a_{11} = -3 + (11-1)(5/2)$ Step 3: Simplify the expression. $a_{11} = -3 + 10(5/2)$ $a_{11} = -3 + 25 = 22$. Final answer: The 11th term of the AP is 22.
• Q: Which term of the AP: $3, 8, 13, 18, \dots$ is $78$? A: Step 1: Identify the given values. First term $a = 3$, common difference $d = 8 - 3 = 5$, and $a_n = 78$. Step 2: Substitute these values into the $n$-th term formula. $78 = 3 + (n-1)5$ Step 3: Solve for $n$. $78 - 3 = (n-1)5$ $75 = (n-1)5$ Divide by 5: $15 = n-1$ $n = 15 + 1 = 16$. Final answer: The 16th term of the given AP is 78.
• Q: How many two-digit numbers are divisible by 3? A: Step 1: Formulate the AP. Two-digit numbers divisible by 3 start from 12 and end at 99. The sequence is: $12, 15, 18, \dots, 99$. Step 2: Identify the variables. First term $a = 12$, common difference $d = 3$, and last term $a_n = 99$. Step 3: Solve for $n$ using the formula $a_n = a + (n-1)d$. $99 = 12 + (n-1)3$ $99 - 12 = (n-1)3$ $87 = (n-1)3$ Divide by 3: $29 = n-1$ $n = 30$. Final answer: There are 30 two-digit numbers divisible by 3.
• Q: Find the 20th term from the last term of the AP: $3, 8, 13, \dots, 253$. A: Step 1: Identify the variables from the AP. First term $a = 3$, common difference $d = 5$, and last term $l = 253$. Step 2: Use the shortcut formula for finding the $n$-th term from the end: $\text{Term from end} = l - (n-1)d$ Step 3: Substitute $n = 20$, $l = 253$, and $d = 5$. $\text{20th term from end} = 253 - (20-1)5$ $= 253 - 19(5)$ $= 253 - 95$ $= 158$. Final answer: The 20th term from the last term is 158.

Frequently Asked Questions