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Q2(iii):
Find the sums given below : (iii) –5 + (–8) + (–11) + . . . + (–230)

Solution :

Given: An arithmetic progression (AP) series: $-5 + (-8) + (-11) + \dots + (-230)$.

To Find: The sum of the given arithmetic series.

Step 1: Identify the parameters of the Arithmetic Progression.

The given series is $-5, -8, -11, \dots, -230$.

Let the first term be $a$. Thus, $a = -5$.

Let the common difference be $d$. The common difference is calculated as the difference between any two consecutive terms:

$d = a_2 - a_1 = (-8) - (-5) = -8 + 5 = -3$.

The last term (or $n^{th}$ term) is $a_n = -230$.

Step 2: Determine the number of terms ($n$).

We use the formula for the $n^{th}$ term of an AP: $a_n = a + (n - 1)d$.

Substituting the known values into the formula:

$-230 = -5 + (n - 1)(-3)$

Add $5$ to both sides of the equation:

$-230 + 5 = (n - 1)(-3)$

$-225 = (n - 1)(-3)$

Divide both sides by $-3$:

$\frac{-225}{-3} = n - 1$

$75 = n - 1$

Add $1$ to both sides:

$n = 76$

[Since there are 76 terms in the series].

Step 3: Calculate the sum of the series ($S_n$).

The formula for the sum of the first $n$ terms of an AP when the last term ($l$) is known is:

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

Where $n = 76$, $a = -5$, and $l = -230$.

Substitute the values into the formula:

$S_{76} = \frac{76}{2}(-5 + (-230))$

Simplify the fraction and the expression inside the parentheses:

$S_{76} = 38(-5 - 230)$

$S_{76} = 38(-235)$

Step 4: Perform the final multiplication.

$38 \times (-235) = -8930$

[Calculation: $38 \times 200 = 7600$; $38 \times 30 = 1140$; $38 \times 5 = 190$; $7600 + 1140 + 190 = 8930$].

Final Answer: The sum of the series is -8930.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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