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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


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