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Q1(iii):
Find the sum of the following APs: (iii) 0.6, 1.7, 2.8, . . ., to 100 terms.

Solution :

Given: An Arithmetic Progression (AP) series: $0.6, 1.7, 2.8, \dots$ and the number of terms $n = 100$.

To Find: The sum of the first $100$ terms of the given AP ($S_{100}$).

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 $a = 0.6$.

The common difference $d$ is calculated by subtracting the first term from the second term:

$d = a_2 - a_1$

$d = 1.7 - 0.6$

$d = 1.1$

[Justification: The common difference $d$ in an AP is constant for any two consecutive terms $a_n - a_{n-1}$.]

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:

  • $n = 100$
  • $a = 0.6$
  • $d = 1.1$

Step 3: Substitute the values into the formula.

$S_{100} = \frac{100}{2} [2(0.6) + (100 - 1)(1.1)]$

Step 4: Perform the arithmetic calculations.

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

$S_{100} = 50 [1.2 + (99)(1.1)]$

[Since $\frac{100}{2} = 50$ and $2 \times 0.6 = 1.2$]

Next, calculate the product inside the bracket:

$99 \times 1.1 = 99 \times \frac{11}{10} = \frac{1089}{10} = 108.9$

Now, add the terms inside the bracket:

$S_{100} = 50 [1.2 + 108.9]$

$S_{100} = 50 [110.1]$

Finally, multiply by 50:

$S_{100} = 50 \times 110.1$

$S_{100} = 5 \times 1101$

$S_{100} = 5505$

Final Answer: The sum of the first 100 terms of the AP is 5505.


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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