If the 3rd and the 9th terms of an AP are and respectively, which term of this AP is zero?

Understand the Question The term of an Arithmetic Progression (AP) is given by the formula: where is the first term and is the common difference. We are given two terms: and . These give a system of two linear equations in terms of and . After solving for and , we set the general term and solve for to determine which term is zero. Step 1 · Find the First Term and Common Difference Using the term formula : For the term (): For the term (): Subtracting equation from equation : Substituting into equation : Step 2 · Find Which Term is Zero Let the term of the AP be , so . Substituting and into : Answer > term Common Mistakes Term Index Offset: Writing instead of (e.g., writing instead of ). Sign Errors in Subtraction: Making mistakes while subtracting negative terms, such as calculating as instead of . Sign Error in Expansion: Misinterpreting as instead of when expanding the bracket.

Arithmetic Progressions | Exercise 5.2

Question 9

If the 3rd and the 9th terms of an AP are $4$ and $-8$ respectively, which term of this AP is zero?

Solution

Understand the Question

The $n^{\text{th}}$ term of an Arithmetic Progression (AP) is given by the formula: $a_n = a + (n - 1)d$ where $a$ is the first term and $d$ is the common difference.
We are given two terms: $a_3 = 4$ and $a_9 = -8$ . These give a system of two linear equations in terms of $a$ and $d$ .
After solving for $a$ and $d$ , we set the general term $a_n = 0$ and solve for $n$ to determine which term is zero.

Step 1 · Find the First Term $a$ and Common Difference $d$

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

For the $3^{\text{rd}}$ term ( $a_3 = 4$ ): $a + 2d = 4 \quad \dots (1)$

For the $9^{\text{th}}$ term ( $a_9 = -8$ ): $a + 8d = -8 \quad \dots (2)$

Subtracting equation $(1)$ from equation $(2)$ :

\begin{aligned} (a + 8d) - (a + 2d) &= -8 - 4 \\[0.6em] 6d &= -12 \\[0.6em] d &= \dfrac{-12}{6} \\[0.6em] d &= -2 \end{aligned}

Substituting $d = -2$ into equation $(1)$ :

\begin{aligned} a + 2(-2) &= 4 \\[0.6em] a - 4 &= 4 \\[0.6em] a &= 4 + 4 \\[0.6em] a &= 8 \end{aligned}

Step 2 · Find Which Term is Zero

Let the $n^{\text{th}}$ term of the AP be $0$ , so $a_n = 0$ .

Substituting $a = 8$ and $d = -2$ into $a_n = a + (n - 1)d$ :

\begin{aligned} 0 &= 8 + (n - 1)(-2) \\[0.6em] 0 &= 8 - 2n + 2 \\[0.6em] 0 &= 10 - 2n \\[0.6em] 2n &= 10 \\[0.6em] n &= \dfrac{10}{2} \\[0.6em] n &= 5 \end{aligned}

Answer

$5^{\text{th}}$ term

Common Mistakes

Term Index Offset: Writing $a_n = a + nd$ instead of $a + (n - 1)d$ (e.g., writing $a_3 = a + 3d$ instead of $a + 2d$ ).
Sign Errors in Subtraction: Making mistakes while subtracting negative terms, such as calculating $-8 - 4$ as $-4$ instead of $-12$ .
Sign Error in Expansion: Misinterpreting $(n - 1)(-2)$ as $-2n - 2$ instead of $-2n + 2$ when expanding the bracket.

Question 9

More questions in Exercise 5.2