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Q3(iii):
For the following APs, write the first term and the common difference: (iii) $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, . . .$

Solution :

Given: An Arithmetic Progression (AP) sequence: $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, . . .$

To find: The first term ($a$) and the common difference ($d$) of the given AP.

Step 1: Identifying the First Term

By definition, an Arithmetic Progression is a sequence of numbers such that the difference between consecutive terms is constant. The first term is denoted by $a$ or $a_1$.

Looking at the sequence: $a_1 = \frac{1}{3}, a_2 = \frac{5}{3}, a_3 = \frac{9}{3}, a_4 = \frac{13}{3}$.

Therefore, the first term $a = \frac{1}{3}$.

Step 2: Defining the Common Difference

The common difference ($d$) of an AP is defined as the difference between any two consecutive terms. Mathematically, it is expressed as:

$d = a_{n} - a_{n-1}$ for $n > 1$.

Specifically, for this sequence:

$d = a_2 - a_1$

Step 3: Calculating the Common Difference

Substitute the values of $a_2$ and $a_1$ into the formula:

$d = \frac{5}{3} - \frac{1}{3}$

[Since the denominators are the same, we subtract the numerators directly]

$d = \frac{5 - 1}{3}$

$d = \frac{4}{3}$

Step 4: Verification (Optional but Recommended)

To ensure the sequence is indeed an AP, we verify the difference between the third and second terms:

$d = a_3 - a_2 = \frac{9}{3} - \frac{5}{3} = \frac{4}{3}$

[Since the difference is consistent, the value of $d$ is confirmed as $\frac{4}{3}$]

Final Answer: The first term $a = \frac{1}{3}$ and the common difference $d = \frac{4}{3}$.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.1


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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