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Q1(iv):
Find the sum of the following APs: (iv) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}$, . . ., to 11 terms.

Solution :

Given: An Arithmetic Progression (AP) series: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$ and the number of terms $n = 11$.

To Find: The sum of the first 11 terms ($S_{11}$) of the given AP.

Step 1: Identify the parameters of the Arithmetic Progression.

The first term ($a$) is the first element of the sequence:
$a = \frac{1}{15}$

The common difference ($d$) is calculated by subtracting the first term from the second term:
$d = a_2 - a_1 = \frac{1}{12} - \frac{1}{15}$

To subtract these fractions, we find the Least Common Multiple (LCM) of the denominators 12 and 15:
Multiples of 12: 12, 24, 36, 48, 60
Multiples of 15: 15, 30, 45, 60
LCM = 60

Converting the fractions:
$d = \frac{1 \times 5}{12 \times 5} - \frac{1 \times 4}{15 \times 4} = \frac{5}{60} - \frac{4}{60} = \frac{1}{60}$

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 $n$ 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 $n = 11$, $a = \frac{1}{15}$, and $d = \frac{1}{60}$:
$S_{11} = \frac{11}{2} \left[ 2\left(\frac{1}{15}\right) + (11 - 1)\left(\frac{1}{60}\right) \right]$

Step 4: Perform the arithmetic calculations.

Simplify the expression inside the brackets:
$S_{11} = \frac{11}{2} \left[ \frac{2}{15} + 10\left(\frac{1}{60}\right) \right]$
$S_{11} = \frac{11}{2} \left[ \frac{2}{15} + \frac{10}{60} \right]$

Simplify the fraction $\frac{10}{60}$ to $\frac{1}{6}$:
$S_{11} = \frac{11}{2} \left[ \frac{2}{15} + \frac{1}{6} \right]$

Find the LCM of 15 and 6 to add the fractions inside the brackets:
Multiples of 15: 15, 30
Multiples of 6: 6, 12, 18, 24, 30
LCM = 30

Convert fractions to have a common denominator of 30:
$\frac{2}{15} = \frac{2 \times 2}{15 \times 2} = \frac{4}{30}$
$\frac{1}{6} = \frac{1 \times 5}{6 \times 5} = \frac{5}{30}$

Add the fractions:
$S_{11} = \frac{11}{2} \left[ \frac{4}{30} + \frac{5}{30} \right] = \frac{11}{2} \left[ \frac{9}{30} \right]$

Step 5: Final multiplication.

$S_{11} = \frac{11}{2} \times \frac{9}{30}$
Simplify $\frac{9}{30}$ by dividing both numerator and denominator by 3:
$\frac{9}{30} = \frac{3}{10}$
$S_{11} = \frac{11}{2} \times \frac{3}{10} = \frac{33}{20}$

Final Answer: The sum of the first 11 terms is $\frac{33}{20}$.


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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