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Q4(xv):
Which of the following are APs ? If they form an AP, find the common difference $d$ and write three more terms. (xv) $1^2, 5^2, 7^2, 73, . . .$

Solution :

Given: A sequence of numbers: $1^2, 5^2, 7^2, 73, \dots$

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

Step 1: Simplifying the terms of the sequence

Let the given sequence be denoted by $a_1, a_2, a_3, a_4, \dots$

Calculating the values of each term:

$a_1 = 1^2 = 1$

$a_2 = 5^2 = 25$

$a_3 = 7^2 = 49$

$a_4 = 73$

The sequence is: $1, 25, 49, 73, \dots$

Step 2: Checking for a common difference

A sequence is an AP if the difference between consecutive terms ($a_{n+1} - a_n$) is constant. This constant is called the common difference $d$.

Calculate the difference between the first and second terms ($d_1$):

$d_1 = a_2 - a_1 = 25 - 1 = 24$

Calculate the difference between the second and third terms ($d_2$):

$d_2 = a_3 - a_2 = 49 - 25 = 24$

Calculate the difference between the third and fourth terms ($d_3$):

$d_3 = a_4 - a_3 = 73 - 49 = 24$

[Since $d_1 = d_2 = d_3 = 24$, the difference between consecutive terms is constant.]

Step 3: Conclusion on AP status

Because the difference between consecutive terms is constant ($d = 24$), the given sequence forms an Arithmetic Progression.

Step 4: Finding the next three terms

To find the next three terms ($a_5, a_6, a_7$), we add the common difference $d = 24$ to the preceding term.

Fifth term ($a_5$):

$a_5 = a_4 + d = 73 + 24 = 97$

Sixth term ($a_6$):

$a_6 = a_5 + d = 97 + 24 = 121$

Seventh term ($a_7$):

$a_7 = a_6 + d = 121 + 24 = 145$

Final Answer: The sequence forms an AP with a common difference $d = 24$. The next three terms are 97, 121, and 145.


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


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