default_background

Find the best tutors and institutes for Class 10 Tuition

Find Best Class 10 Tuition

Please select a Category.

Please select a Locality.

No matching category found.

No matching Locality found.

Q4(v):
Which of the following are APs ? If they form an AP, find the common difference $d$ and write three more terms. (v) $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, . . .$

Solution :

Given: A sequence of numbers: $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots$

To find: Determine if the sequence forms an Arithmetic Progression (AP). If it does, find the common difference $d$ and write the next three terms.

Step 1: Definition of an Arithmetic Progression
An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference ($d$).
Let the terms be $a_1, a_2, a_3, a_4, \dots$
Here, $a_1 = 3$, $a_2 = 3+\sqrt{2}$, $a_3 = 3+2\sqrt{2}$, $a_4 = 3+3\sqrt{2}$.

Step 2: Calculating the differences between consecutive terms
We calculate the difference $d_n = a_{n+1} - a_n$ for consecutive terms:

Difference 1 ($d_1$):
$d_1 = a_2 - a_1 = (3 + \sqrt{2}) - 3$
$d_1 = 3 - 3 + \sqrt{2} = \sqrt{2}$

Difference 2 ($d_2$):
$d_2 = a_3 - a_2 = (3 + 2\sqrt{2}) - (3 + \sqrt{2})$
$d_2 = 3 + 2\sqrt{2} - 3 - \sqrt{2}$
$d_2 = (3 - 3) + (2\sqrt{2} - \sqrt{2}) = \sqrt{2}$

Difference 3 ($d_3$):
$d_3 = a_4 - a_3 = (3 + 3\sqrt{2}) - (3 + 2\sqrt{2})$
$d_3 = 3 + 3\sqrt{2} - 3 - 2\sqrt{2}$
$d_3 = (3 - 3) + (3\sqrt{2} - 2\sqrt{2}) = \sqrt{2}$

Step 3: Verification
Since $d_1 = d_2 = d_3 = \sqrt{2}$, the difference between consecutive terms is constant. Therefore, the given sequence is an Arithmetic Progression with common difference $d = \sqrt{2}$.

Step 4: Finding the next three terms
To find the next terms, we add the common difference $d = \sqrt{2}$ to the last known term ($a_4 = 3 + 3\sqrt{2}$):

Fifth term ($a_5$):
$a_5 = a_4 + d = (3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2}$

Sixth term ($a_6$):
$a_6 = a_5 + d = (3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2}$

Seventh term ($a_7$):
$a_7 = a_6 + d = (3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2}$

Final Answer: The sequence forms an AP with common difference $d = \sqrt{2}$. The next three terms are $3+4\sqrt{2}, 3+5\sqrt{2},$ and $3+6\sqrt{2}$.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.1


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


Other Subjects in CBSE - Class 10

Worksheet Icon

Download free CBSE - Class 10 Mathematics Arithmetic Progression EXERCISE 5.1 worksheets

Download Now

Find Best Class 10 Tuition ?

Find Now »