NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.3

Question 1

Formula used: For an AP with first term \(a\), common difference \(d\) and \(n\) terms, the sum of the first \(n\) terms is

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

We first identify \(a\), \(d\) and \(n\) for each AP, then apply the formula.

(i) \(2, 7, 12, \ldots\), to 10 terms

First term: \(a = 2\)

Common difference: \(d = 7 - 2 = 5\)

Number of terms: \(n = 10\)

Sum:

\[S_{10} = \frac{10}{2} \{2 \cdot 2 + (10 - 1) \cdot 5\}\]

\[S_{10} = 5 \{4 + 9 \cdot 5\} = 5 \{4 + 45\} = 5 \cdot 49 = 245\]

So, sum = 245.

(ii) \(-37, -33, -29, \ldots\), to 12 terms

First term: \(a = -37\)

Common difference: \(d = -33 - (-37) = 4\)

Number of terms: \(n = 12\)

Sum:

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

\[S_{12} = 6 \{-74 + 11 \cdot 4\} = 6 \{-74 + 44\} = 6 \cdot (-30) = -180\]

So, sum = -180.

(iii) \(0.6, 1.7, 2.8, \ldots\), to 100 terms

First term: \(a = 0.6\)

Common difference: \(d = 1.7 - 0.6 = 1.1\)

Number of terms: \(n = 100\)

Sum:

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

\[S_{100} = 50 \{1.2 + 99 \cdot 1.1\}\]

\[99 \cdot 1.1 = 108.9\]

So,

\[S_{100} = 50 \cdot (1.2 + 108.9) = 50 \cdot 110.1 = 5505\]

So, sum = 5505.

(iv) \(\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}, \ldots\), to 11 terms

First term: \(a = \dfrac{1}{15}\)

Common difference:

\[d = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}\]

Number of terms: \(n = 11\)

Sum:

\[S_{11} = \frac{11}{2} \left\{2 \cdot \frac{1}{15} + (11 - 1) \cdot \frac{1}{60} \right\}\]

\[2 \cdot \frac{1}{15} = \frac{2}{15} = \frac{8}{60}\]

\[10 \cdot \frac{1}{60} = \frac{10}{60} = \frac{1}{6} = \frac{10}{60}\]

So inside the brackets:

\[\frac{8}{60} + \frac{10}{60} = \frac{18}{60} = \frac{3}{10}\]

Thus,

\[S_{11} = \frac{11}{2} \cdot \frac{3}{10} = \frac{33}{20}\]

So, sum = \(\dfrac{33}{20}\).