Check whether is a term of the AP:

Understand the Question The term of an Arithmetic Progression is given by the formula , where is the first term and is the common difference. The term number represents the position of a term in the sequence and therefore must always be a positive integer (natural number: ). To check if is a term of the AP, we assume and solve for . If is a positive integer, it is a term; if is a fraction or negative, it is not. Step 1 · Find the First Term and Common Difference Given AP: Step 2 · Set up the Term Formula and Solve for Assume is the term () of the AP. Using the formula Step 3 · Check if is a Positive Integer Since must be a positive integer and is a fraction, cannot be a term of this AP. Answer > No, is not a term of the given AP. Common Mistakes Common Difference Sign Error: Calculating instead of . Since the sequence decreases, must be negative. Condition for : Forgetting that represents the index/count of a term, which must strictly be a positive integer (). A fraction or negative value means the term does not exist in the AP.

Arithmetic Progressions | Exercise 5.2

Question 6

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

Solution

Understand the Question

The $n^{\text{th}}$ term of an Arithmetic Progression is given by the formula $a_n = a + (n - 1)d$ , where $a$ is the first term and $d$ is the common difference.
The term number $n$ represents the position of a term in the sequence and therefore must always be a positive integer (natural number: $1, 2, 3, \dots$ ).
To check if $-150$ is a term of the AP, we assume $a_n = -150$ and solve for $n$ . If $n$ is a positive integer, it is a term; if $n$ is a fraction or negative, it is not.

Step 1 · Find the First Term and Common Difference

Given AP: $11, 8, 5, 2, \dots$ Diagram 1

\begin{aligned} a &= 11 \\[0.6em] d &= 8 - 11 \\[0.6em] &= -3 \end{aligned}

Step 2 · Set up the $n^{\text{th}}$ Term Formula and Solve for $n$

Assume $-150$ is the $n^{\text{th}}$ term ( $a_n$ ) of the AP.

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

\begin{aligned} -150 &= 11 + (n - 1)(-3) \\[0.6em] -150 - 11 &= (n - 1)(-3) \\[0.6em] -161 &= -3(n - 1) \\[0.6em] \dfrac{-161}{-3} &= n - 1 \\[0.6em] \dfrac{161}{3} &= n - 1 \\[0.6em] n &= \dfrac{161}{3} + 1 \\[0.6em] n &= \dfrac{161 + 3}{3} \\[0.6em] n &= \dfrac{164}{3} \end{aligned}

Step 3 · Check if $n$ is a Positive Integer

\dfrac{164}{3} = 54.66\dots

Since $n$ must be a positive integer and $\dfrac{164}{3}$ is a fraction, $-150$ cannot be a term of this AP.

Answer

No, $-150$ is not a term of the given AP.

Common Mistakes

Common Difference Sign Error: Calculating $d = 11 - 8 = 3$ instead of $d = a_2 - a_1 = 8 - 11 = -3$ . Since the sequence decreases, $d$ must be negative.
Condition for $n$ : Forgetting that $n$ represents the index/count of a term, which must strictly be a positive integer ( $n \in \mathbb{N}$ ). A fraction or negative value means the term does not exist in the AP.

Question 6

More questions in Exercise 5.2