The nth Term of an AP and Board Problems for CBSE Class 10

Once you can identify an Arithmetic Progression and compute its common difference, the next logical question is: How can we find the 50th term, the 100th term, or any arbitrary term without writing out every intermediate number?

The $n$-th term formula (also called the general term) provides an exact algebraic equation that links the position of any term to its numerical value. In CBSE Class 10 Mathematics, questions involving the $n$-th term appear in every board examination, carrying anywhere from 1 to 4 marks.

What You Will Learn

• Algebraic derivation of the general term formula: $a_n = a + (n - 1)d$
• Meaning and domain of the four variables ($a_n, a, n, d$)
• Finding the $n$-th term from the end of a finite AP ($l - (n - 1)d$)
• Determining whether a given number belongs to an AP (the natural number test for $n$)
• Setting up and solving simultaneous linear equations to find $a$ and $d$
• High-yield CBSE board exam questions and error-prevention tips

1. Derivation of the Formula for the nth Term

Let an AP have first term $a$ and common difference $d$. The successive terms are formed as follows:

• 1st term: $a_1 = a = a + (1 - 1)d$
• 2nd term: $a_2 = a + d = a + (2 - 1)d$
• 3rd term: $a_3 = a_2 + d = (a + d) + d = a + 2d = a + (3 - 1)d$
• 4th term: $a_4 = a_3 + d = (a + 2d) + d = a + 3d = a + (4 - 1)d$

Observing the clear pattern, the coefficient of $d$ is always one less than the term index $n$. Therefore:

The General Term Formula

The $n$-th term $a_n$ of an Arithmetic Progression with first term $a$ and common difference $d$ is given by: $a_n = a + (n - 1)d$

If the AP is finite with total $m$ terms, then $a_m$ represents the last term, frequently denoted by the letter $l$: $l = a + (n - 1)d$

Important: <u>The term index $n$ represents a counting position. Therefore, $n$ must ALWAYS be a positive integer (natural number: $n \in \{1, 2, 3, 4, \dots\}$). A term number can never be a fraction, a decimal, or negative!</u>

2. Finding the nth Term from the End of an AP

In board exams, questions often ask: "Find the 10th term from the last term of the AP..."

Method 1: The Direct Reverse Formula

If a finite AP has last term $l$ and common difference $d$, moving backwards from the end means the common difference becomes $-d$. Therefore:

The $n$-th term from the end is: $a_n' = l - (n - 1)d$

Method 2: Reverse the Sequence

Simply write the AP in reverse order:

• The last term becomes the new first term: $a' = l$.
• The common difference reverses sign: $d' = -d$.
• Find the $n$-th term using the standard formula $a_n = a' + (n - 1)d'$.

3. Solved Board Examination Problems

Solved Example 1: Finding a High-Index Term

Problem: Find the 30th term of the AP: $10, 7, 4, \dots$.

Solution:

1. First term: $a = 10$.
2. Common difference: $d = 7 - 10 = -3$.
3. Term index: $n = 30$.
4. Apply formula: $a_{30} = a + (30 - 1)d = 10 + 29(-3)$ $a_{30} = 10 - 87 = -77$
5. Therefore, <u>the 30th term is $-77$</u>.

Solved Example 2: Testing Whether a Number is a Term of an AP

Problem: Check whether $-150$ is a term of the AP: $11, 8, 5, 2, \dots$.

Solution:

1. Here $a = 11$, and $d = 8 - 11 = -3$.
2. Suppose $-150$ is the $n$-th term of this AP: $a_n = a + (n - 1)d = -150$
3. Substitute $a$ and $d$: $11 + (n - 1)(-3) = -150$ $(n - 1)(-3) = -150 - 11 = -161$ $n - 1 = \frac{-161}{-3} = \frac{161}{3}$ $n = \frac{161}{3} + 1 = \frac{164}{3} = 54\frac{2}{3}$
4. Since $n$ is a fraction ($54.67$) and not a positive integer, <u>$-150$ is NOT a term of this AP</u>.

Solved Example 3: Finding $a$ and $d$ from Given Conditions

Problem: An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:

1. Given: Total terms $n = 50$, so the last term is the 50th term ($a_{50} = 106$). Third term $a_3 = 12$.
2. Express in terms of $a$ and $d$: $a_3 = a + 2d = 12 \quad \text{--- (1)}$ $a_{50} = a + 49d = 106 \quad \text{--- (2)}$
3. Subtract Equation (1) from Equation (2): $(a + 49d) - (a + 2d) = 106 - 12$ $47d = 94 \implies d = \frac{94}{47} = 2$
4. Substitute $d = 2$ into (1): $a + 2(2) = 12 \implies a + 4 = 12 \implies a = 8$
5. Now calculate the 29th term: $a_{29} = a + (29 - 1)d = 8 + 28(2) = 8 + 56 = 64$
6. Therefore, <u>the 29th term is $64$</u>.

Solved Example 4: Finding a Term from the End (CBSE PYQ)

Problem: Find the 11th term from the last term of the AP: $10, 7, 4, \dots, -62$.

Solution:

1. Here $a = 10$, common difference $d = 7 - 10 = -3$, and last term $l = -62$.
2. We want the 11th term from the end ($n = 11$): $a_{11}' = l - (n - 1)d$ $a_{11}' = -62 - (11 - 1)(-3)$ $a_{11}' = -62 - (10)(-3) = -62 + 30 = -32$
3. Therefore, <u>the 11th term from the end is $-32$</u>.

4. Summary and Examination Tips

Exam Tip: Whenever you solve for $n$ in a "Check whether X is a term" question, always conclude with: "Since $n$ is not a natural number (positive integer), X cannot be a term of the AP."

Common Mistake: Writing $(n - 1)d$ as $nd - 1$. The expression is $n - 1$ grouped in brackets, multiplied by $d$: $(n - 1) \times d$!