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Q3(ii):
In the following APs, find the missing terms in the boxes : (ii) ☐, 13, ☐, 3

Solution :

Given: An Arithmetic Progression (AP) with missing terms represented as: $\square, 13, \square, 3$.

To find: The missing terms in the first and third positions of the sequence.

Step 1: Defining the variables of the Arithmetic Progression

Let the terms of the Arithmetic Progression be denoted by $a_1, a_2, a_3, a_4$.
The general form of an AP is given by $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.
From the given sequence:
$a_1 = \square$
$a_2 = 13$
$a_3 = \square$
$a_4 = 3$

Step 2: Formulating the equations based on the general term formula

Using the formula $a_n = a + (n-1)d$:

For the second term ($n=2$):
$a_2 = a + (2-1)d = a + d = 13$ --- (Equation 1)

For the fourth term ($n=4$):
$a_4 = a + (4-1)d = a + 3d = 3$ --- (Equation 2)

Step 3: Solving the system of linear equations

To find the common difference $d$, subtract Equation 1 from Equation 2:
$(a + 3d) - (a + d) = 3 - 13$
$a - a + 3d - d = -10$
$2d = -10$
$d = \frac{-10}{2}$
$d = -5$

Now, substitute the value of $d$ into Equation 1 to find the first term $a$:
$a + (-5) = 13$
$a = 13 + 5$
$a = 18$

Step 4: Calculating the missing terms

The first term ($a_1$) is $a = 18$.

The third term ($a_3$) is calculated as:
$a_3 = a + 2d$
$a_3 = 18 + 2(-5)$ [Substituting $a=18$ and $d=-5$]
$a_3 = 18 - 10$
$a_3 = 8$

Verification:
The sequence is $18, 13, 8, 3$.
Common difference check: $13 - 18 = -5$; $8 - 13 = -5$; $3 - 8 = -5$.
Since the common difference is constant, the values are correct.

Final Answer: The missing terms are 18 and 8. The complete AP is 18, 13, 8, 3.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.2


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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