Arithmetic Progressions | Exercise 5.2

Question 5

Find the number of terms in each of the following APs:

(i) $7, 13, 19, \dots, 205$

(ii) $18, 15\dfrac{1}{2}, 13, \dots, -47$

Solution

Understand the Question

An Arithmetic Progression (AP) has a first term $a$ , a common difference $d$ , and an $n^{\text{th}}$ term $a_n$ .
To find the total number of terms $n$ in a finite AP when the last term is known, we use the general formula: $a_n = a + (n - 1)d$
We identify $a$ , calculate $d = a_2 - a_1$ , substitute $a_n$ , and solve the linear equation for $n$ .

(i) Find the number of terms in the AP: $7, 13, 19, \dots, 205$

Step 1 · Find the Number of Terms

For the given AP $7, 13, 19, \dots, 205$ :

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

\begin{aligned} 205 &= 7 + (n - 1)6 \\[0.6em] 205 - 7 &= (n - 1)6 \\[0.6em] 198 &= (n - 1)6 \\[0.6em] \dfrac{198}{6} &= n - 1 \\[0.6em] 33 &= n - 1 \\[0.6em] n &= 33 + 1 \\[0.6em] n &= 34 \end{aligned}

Answer

(i) $34$

(ii) Find the number of terms in the AP: $18, 15\dfrac{1}{2}, 13, \dots, -47$

Step 1 · Find the Number of Terms

For the given AP $18, 15\dfrac{1}{2}, 13, \dots, -47$ :

\begin{aligned} d &= \dfrac{31}{2} - 18 \\[0.6em] &= \dfrac{31 - 36}{2} \\[0.6em] &= -\dfrac{5}{2} = -2.5 \end{aligned}

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

\begin{aligned} -47 &= 18 + (n - 1)(-2.5) \\[0.6em] -47 - 18 &= (n - 1)(-2.5) \\[0.6em] -65 &= (n - 1)(-2.5) \\[0.6em] \dfrac{-65}{-2.5} &= n - 1 \\[0.6em] 26 &= n - 1 \\[0.6em] n &= 26 + 1 \\[0.6em] n &= 27 \end{aligned}

Answer

(ii) $27$

Common Mistakes

Sign Error in Common Difference: For a decreasing AP, $d$ is negative ( $d = 15\dfrac{1}{2} - 18 = -\dfrac{5}{2}$ ). Omitting the negative sign gives a negative or non-integer value for $n$ .
Forgetting to Add 1: Stopping at $n - 1 = 33$ instead of adding $1$ to find $n = 34$ .
Fraction Calculation Errors: Incorrectly converting mixed fraction $15\dfrac{1}{2}$ to $\dfrac{31}{2}$ or miscalculating $\dfrac{31}{2} - 18$ .

More questions in Exercise 5.2