Arithmetic Progressions | Exercise 5.2

Question 3

In the following APs, find the missing terms in the boxes :

Solution

Understand the Question

For an Arithmetic Progression (AP), the $n$ -th term is given by the formula: $a_n = a + (n - 1)d$ where $a$ is the first term, $d$ is the common difference, and $n$ is the position of the term.
To find the missing terms in each AP:
1. Identify the given terms and their respective positions ( $n$ ).
2. Set up algebraic equations in terms of $a$ and $d$ .
3. Solve for $a$ and $d$ , then calculate the missing terms.

(i) $2, \boxed{\phantom{x}}, 26$

Step 1 · Find Common Difference and Missing Term

Given $a_1 = 2$ and $a_3 = 26$ .

\begin{aligned} a_3 &= a_1 + 2d \\ 26 &= 2 + 2d \\ 24 &= 2d \\ d &= 12 \end{aligned}

Now, calculate the missing term $a_2$ :

\begin{aligned} a_2 &= a_1 + d \\ &= 2 + 12 \\ &= 14 \end{aligned}

Answer

(i) $14$

(ii) $\boxed{\phantom{x}}, 13, \boxed{\phantom{x}}, 3$

Step 1 · Find First Term and Common Difference

Given $a_2 = 13$ and $a_4 = 3$ .

\begin{aligned} a_2 &= a_1 + d \implies 13 = a_1 + d \quad \dots (1) \\ a_4 &= a_1 + 3d \implies 3 = a_1 + 3d \quad \dots (2) \end{aligned}

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

\begin{aligned} (a_1 + 3d) - (a_1 + d) &= 3 - 13 \\ 2d &= -10 \\ d &= -5 \end{aligned}

Substitute $d = -5$ into equation $(1)$ to find $a_1$ :

\begin{aligned} 13 &= a_1 + (-5) \\ a_1 &= 13 + 5 \\ a_1 &= 18 \end{aligned}

Step 2 · Find the Third Term

Calculate $a_3$ :

\begin{aligned} a_3 &= a_2 + d \\ &= 13 + (-5) \\ &= 8 \end{aligned}

Answer

(ii) $18, 8$

(iii) $5, \boxed{\phantom{x}}, \boxed{\phantom{x}}, 9\dfrac{1}{2}$

Step 1 · Find Common Difference

Given $a_1 = 5$ and $a_4 = 9\dfrac{1}{2} = \dfrac{19}{2}$ .

\begin{aligned} a_4 &= a_1 + 3d \\[0.6em] \dfrac{19}{2} &= 5 + 3d \\[0.6em] \dfrac{19}{2} - \dfrac{10}{2} &= 3d \\[0.6em] \dfrac{9}{2} &= 3d \\[0.6em] d &= \dfrac{9}{2 \times 3} \\[0.6em] d &= \dfrac{3}{2} \end{aligned}

Step 2 · Find Missing Terms

Calculate $a_2$ and $a_3$ :

\begin{aligned} a_2 &= a_1 + d \\[0.6em] &= 5 + \dfrac{3}{2} \\[0.6em] &= \dfrac{10}{2} + \dfrac{3}{2} \\[0.6em] &= \dfrac{13}{2} = 6\dfrac{1}{2} \end{aligned}

\begin{aligned} a_3 &= a_2 + d \\[0.6em] &= \dfrac{13}{2} + \dfrac{3}{2} \\[0.6em] &= \dfrac{16}{2} \\[0.6em] &= 8 \end{aligned}

Answer

(iii) $\dfrac{13}{2}, 8$

(iv) $-4, \boxed{\phantom{x}}, \boxed{\phantom{x}}, \boxed{\phantom{x}}, \boxed{\phantom{x}}, 6$

Step 1 · Find Common Difference

Given $a_1 = -4$ and $a_6 = 6$ .

\begin{aligned} a_6 &= a_1 + 5d \\ 6 &= -4 + 5d \\ 6 + 4 &= 5d \\ 10 &= 5d \\ d &= 2 \end{aligned}

Step 2 · Find Missing Terms

Calculate $a_2, a_3, a_4,$ and $a_5$ :

\begin{aligned} a_2 &= a_1 + d = -4 + 2 = -2 \\[0.6em] a_3 &= a_2 + d = -2 + 2 = 0 \\[0.6em] a_4 &= a_3 + d = 0 + 2 = 2 \\[0.6em] a_5 &= a_4 + d = 2 + 2 = 4 \end{aligned}

Answer

(iv) $-2, 0, 2, 4$

(v) $\boxed{\phantom{x}}, 38, \boxed{\phantom{x}}, \boxed{\phantom{x}}, \boxed{\phantom{x}}, -22$

Step 1 · Find First Term and Common Difference

Given $a_2 = 38$ and $a_6 = -22$ .

\begin{aligned} a_2 &= a_1 + d \implies 38 = a_1 + d \quad \dots (1) \\ a_6 &= a_1 + 5d \implies -22 = a_1 + 5d \quad \dots (2) \end{aligned}

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

\begin{aligned} (a_1 + 5d) - (a_1 + d) &= -22 - 38 \\ 4d &= -60 \\ d &= -15 \end{aligned}

Substitute $d = -15$ into equation $(1)$ to find $a_1$ :

\begin{aligned} 38 &= a_1 + (-15) \\ a_1 &= 38 + 15 \\ a_1 &= 53 \end{aligned}

Step 2 · Find Remaining Missing Terms

Calculate $a_3, a_4,$ and $a_5$ :

\begin{aligned} a_3 &= a_2 + d = 38 + (-15) = 23 \\[0.6em] a_4 &= a_3 + d = 23 + (-15) = 8 \\[0.6em] a_5 &= a_4 + d = 8 + (-15) = -7 \end{aligned}

Answer

(v) $53, 23, 8, -7$

Common Mistakes

Counting Total Terms ( $n$ ): Misidentifying the position of given terms by skipping empty boxes (e.g., mistaking the $6^{\text{th}}$ term for the $5^{\text{th}}$ term).
Negative Common Difference ( $d$ ): Making sign errors during subtraction when $d$ is negative (e.g., $38 + (-15)$ becoming $53$ instead of $23$ ).
Fraction Conversion Errors: Incorrectly converting mixed numbers to improper fractions (e.g., writing $9\dfrac{1}{2}$ as $\dfrac{18}{2}$ instead of $\dfrac{19}{2}$ ).

More questions in Exercise 5.2