Find the and terms of the AP:

Understand the Question An Arithmetic Progression (AP) is a sequence of numbers where each term is obtained by adding a fixed number (the common difference, ) to the preceding term. For the given AP , the first term is and the common difference is . The term of an AP is given by the formula: We substitute and into the formula to find the respective terms. Step 1 · Find First Term and Common Difference Given AP: First term: Common difference: Step 2 · Calculate the 10th Term Using the term formula: Substitute , , and : Step 3 · Calculate the 26th Term Substitute , , and into the formula: Answer > The term is and the term is . Common Mistakes Using Instead of : Writing the formula as instead of . Sign of Common Difference: Subtracting in the wrong order () instead of (). Order of Operations: Adding and before multiplying by (e.g., calculating instead of ).

Predicting What Comes Next: Sequences and Progressions | Exercise 8.2

Question 1

Find the $10^{\text{th}}$ and $26^{\text{th}}$ terms of the AP: $3, 8, 13, 18, \dots$

Solution

Understand the Question

An Arithmetic Progression (AP) is a sequence of numbers where each term is obtained by adding a fixed number (the common difference, $d$ ) to the preceding term.
For the given AP $3, 8, 13, 18, \dots$ , the first term is $a = 3$ and the common difference is $d = 8 - 3 = 5$ .
The $n^{\text{th}}$ term of an AP is given by the formula: $t_n = a + (n - 1)d$
We substitute $n = 10$ and $n = 26$ into the formula to find the respective terms.

Step 1 · Find First Term and Common Difference

Given AP: $3, 8, 13, 18, \dots$

First term: $a = 3$

Common difference: $d = 8 - 3 = 5$ Diagram 1

Step 2 · Calculate the 10th Term

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

Substitute $a = 3$ , $d = 5$ , and $n = 10$ :

\begin{aligned} t_{10} &= 3 + (10 - 1) \times 5 \\ &= 3 + (9) \times 5 \\ &= 3 + 45 \\ &= 48 \end{aligned}

Step 3 · Calculate the 26th Term

Substitute $a = 3$ , $d = 5$ , and $n = 26$ into the formula:

\begin{aligned} t_{26} &= 3 + (26 - 1) \times 5 \\ &= 3 + (25) \times 5 \\ &= 3 + 125 \\ &= 128 \end{aligned}

Answer

The $10^{\text{th}}$ term is $48$ and the $26^{\text{th}}$ term is $128$ .

Common Mistakes

Using $n$ Instead of $(n-1)$ : Writing the formula as $t_n = a + nd$ instead of $t_n = a + (n - 1)d$ .
Sign of Common Difference: Subtracting in the wrong order ( $3 - 8 = -5$ ) instead of ( $a_2 - a_1 = 8 - 3 = 5$ ).
Order of Operations: Adding $a$ and $(n-1)$ before multiplying by $d$ (e.g., calculating $(3 + 9) \times 5$ instead of $3 + (9 \times 5)$ ).

Question 1

More questions in Exercise 8.2