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Q3(iii):
In an AP: (iii) given $a_{12} = 37, d = 3$, find $a$ and $S_{12}$.

Solution :

Given:

In an Arithmetic Progression (AP):

1. The 12th term, $a_{12} = 37$

2. The common difference, $d = 3$

To find:

1. The first term, $a$

2. The sum of the first 12 terms, $S_{12}$

Step 1: Finding the first term ($a$)

We use the general formula for the $n^{th}$ term of an Arithmetic Progression:

$a_n = a + (n - 1)d$

Substituting the given values $n = 12$, $a_{12} = 37$, and $d = 3$ into the formula:

$37 = a + (12 - 1) \times 3$

$37 = a + (11) \times 3$

$37 = a + 33$

Subtracting 33 from both sides to isolate $a$:

$a = 37 - 33$

$a = 4$

Step 2: Finding the sum of the first 12 terms ($S_{12}$)

We use the formula for the sum of the first $n$ terms of an AP when the first and last terms are known:

$S_n = \frac{n}{2} (a + a_n)$

Here, $n = 12$, $a = 4$, and $a_{12} = 37$. Substituting these values:

$S_{12} = \frac{12}{2} (4 + 37)$

Performing the division and addition inside the parentheses:

$S_{12} = 6 \times (41)$

Calculating the final product:

$S_{12} = 246$

Summary of Results:

The first term $a$ is calculated to be $4$.

The sum of the first 12 terms $S_{12}$ is calculated to be $246$.

Final Answer: $a = 4$ and $S_{12} = 246$


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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