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Q15:
For what value of $n$, are the $n$th terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

Solution :

Given:

Two Arithmetic Progressions (APs):

AP 1: $63, 65, 67, \dots$

AP 2: $3, 10, 17, \dots$

To Find:

The value of $n$ for which the $n$th term of AP 1 is equal to the $n$th term of AP 2.

Step 1: Identify the parameters of the first AP.

For the first AP: $63, 65, 67, \dots$

Let the first term be $a_1$ and the common difference be $d_1$.

$a_1 = 63$

$d_1 = 65 - 63 = 2$

The formula for the $n$th term of an AP is $a_n = a + (n - 1)d$.

Therefore, the $n$th term of the first AP ($A_n$) is:

$A_n = 63 + (n - 1)2$

$A_n = 63 + 2n - 2$

$A_n = 61 + 2n$ --- (Equation 1)

Step 2: Identify the parameters of the second AP.

For the second AP: $3, 10, 17, \dots$

Let the first term be $a_2$ and the common difference be $d_2$.

$a_2 = 3$

$d_2 = 10 - 3 = 7$

Therefore, the $n$th term of the second AP ($B_n$) is:

$B_n = 3 + (n - 1)7$

$B_n = 3 + 7n - 7$

$B_n = 7n - 4$ --- (Equation 2)

Step 3: Equate the $n$th terms and solve for $n$.

According to the problem, the $n$th terms are equal, so $A_n = B_n$.

Substituting Equation 1 and Equation 2:

$61 + 2n = 7n - 4$

Rearranging the terms to isolate $n$:

$61 + 4 = 7n - 2n$ [Transposing $-4$ to the left and $2n$ to the right]

$65 = 5n$

$n = \frac{65}{5}$ [Dividing both sides by 5]

$n = 13$

Step 4: Verification (Optional but recommended).

For $n = 13$ in AP 1: $A_{13} = 63 + (13 - 1)2 = 63 + 12(2) = 63 + 24 = 87$.

For $n = 13$ in AP 2: $B_{13} = 3 + (13 - 1)7 = 3 + 12(7) = 3 + 84 = 87$.

Since $87 = 87$, the value is verified.

Final Answer: The value of $n$ for which the $n$th terms of the two APs are equal is 13.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.2


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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