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Q1(ii):
Find the sum of the following APs: (ii) –37, –33, –29, . . ., to 12 terms.

Solution :

Given: An Arithmetic Progression (AP) series: $-37, -33, -29, \dots$ and the number of terms to be summed, $n = 12$.

To Find: The sum of the first $12$ terms of the given AP ($S_{12}$).

Step 1: Identify the parameters of the Arithmetic Progression.

An Arithmetic Progression is defined by its first term ($a$) and its common difference ($d$).

The first term is given by the first element of the sequence:

$a = -37$

The common difference ($d$) is calculated by subtracting any term from the term that follows it:

$d = a_2 - a_1$

$d = -33 - (-37)$

$d = -33 + 37$

$d = 4$

Step 2: State the formula for the sum of the first $n$ terms of an AP.

The sum of the first $n$ terms of an AP is given by the formula:

$S_n = \frac{n}{2} [2a + (n - 1)d]$

[Where $S_n$ is the sum, $n$ is the number of terms, $a$ is the first term, and $d$ is the common difference.]

Step 3: Substitute the known values into the formula.

Given $n = 12$, $a = -37$, and $d = 4$:

$S_{12} = \frac{12}{2} [2(-37) + (12 - 1)(4)]$

Step 4: Perform the arithmetic calculations.

First, simplify the fraction and the terms inside the brackets:

$S_{12} = 6 [2(-37) + (11)(4)]$

[Since $\frac{12}{2} = 6$ and $12 - 1 = 11$]

Next, calculate the products inside the brackets:

$S_{12} = 6 [-74 + 44]$

[Since $2 \times -37 = -74$ and $11 \times 4 = 44$]

Perform the addition inside the brackets:

$S_{12} = 6 [-30]$

[Since $-74 + 44 = -30$]

Finally, multiply the result by 6:

$S_{12} = -180$

Final Answer: The sum of the first 12 terms of the AP is -180.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


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