# Ex.5.3 Q1 Arithmetic progressions Solutions - NCERT Maths Class 10

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## Question

Find the sum of the following APs.

(i) $$2, 7, 12 ,\,\dots,$$ to $$10$$ terms.

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

(iii) $$0.6, 1.7, 2.8 ,\,\dots,$$ to $$100$$ terms

(iv) \begin{align}\frac{1}{{15}},\frac{1}{{12}},\frac{1}{{10}},\end{align}........., to $$11$$ terms

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 1

## Text Solution

(i) $$2, 7, 12 ,\,\dots,$$ to $$10$$ terms.

What is Known?

The AP $$2,7,12,\, \dots$$

#### What is Unknown?

Sum upto $$10$$ terms of the AP.

#### Reasoning:

Sum of the first $$n$$ terms of an AP is given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

#### Given,

• First term, \begin{align}{a = 2}\end{align}
• Common Difference, $$d = 7 - 2 = 5$$
• Number of Terms, \begin{align}n = 10\end{align}

We know that Sum up to $$n^\rm{th}$$ term of AP,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{10}} &= \frac{{10}}{2}\left[ {2\left( 2 \right) + \left( {10 - 1} \right)5} \right]\\ &= 5\left[ {4 + 9 \times 5 } \right]\\ &= 5\left[ {4 + 45} \right]\\ &= 5 \times 49 \\&= 245\end{align}

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

#### What is Known?

The AP $$-37, -33, -29, \dots$$

#### What is Unknown?

Sum up to $$12$$ terms

#### Reasoning:

Sum of the first $$n$$ terms of an AP is given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Given,

• First term, \begin{align}a = -37\end{align}
• Common Difference, $$d = ( - 33) - ( - 37) = 4$$
• Number of Terms, \begin{align}n = 12\end{align}

We know that Sum upto  $$n^\rm{th}$$ term of AP,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{12}} &= \frac{{12}}{2}\left[ {2\left( { - 37} \right) + \left( {12 - 1} \right)4} \right]\\ &= 6\left[ { - 74 + 11 \times 4} \right]\\& = 6\left[ { - 74 + 44} \right]\\& = 6\times\left( { - 30} \right)\\& = - 180\end{align}

(iii) $$0.6, 1.7, 2.8 ,\,\dots,$$ to $$100$$ terms

What is Known?

The AP $$0.6, 1.7, 2.8 ,\,\dots,$$

#### What is Unknown?

Sum up to $$100$$ terms of the AP.

#### Reasoning:

Sum of the first $$n$$ terms of an AP is given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

#### Steps:

Given,

• First term, \begin{align}a = 0.6\end{align}
• Common difference, $$d = 1.7 - 0.6 = 1.1$$
• Number of Terms, \begin{align}n = 100\end{align}

We know that Sum up to $$n^\rm{th}$$ term of AP,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{100}} &= \frac{{100}}{2}\left[ {2 \times 0.6 + \left( {100 - 1} \right)1.1} \right]\\ &= 50\left[ {1.2 + {99} \times {1.1} } \right]\\ &= 50\left[ {1.2 + 108.9} \right]\\ &= 50\left[ {110.1} \right]\\ &= 5505\end{align}

(iv) \begin{align}\frac{1}{{15}},\frac{1}{{12}},\frac{1}{{10}},\end{align}........., to $$11$$ terms

#### What is Known?

The AP \begin{align}\frac{1}{{15}},\frac{1}{{12}},\frac{1}{{10}}, \end{align}

#### What is Unknown?

Sum up to $$11$$ terms of the AP

#### Reasoning:

Sum of the first $$n$$ terms of an AP is given by $${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Given,

• First term, \begin{align}a = \frac{1}{{15}}\end{align}
• Common difference, \begin{align}d = \frac{1}{{12}} - \frac{1}{{15}} = \frac{1}{{60}}\end{align}
• Number of Terms, \begin{align}n=11\end{align}

We know that Sum up to $$n^\rm{th}$$ term of AP,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{11}} &= \frac{{11}}{2}\left[ {2 \times \frac{1}{{15}} + \left( {11 - 1} \right)\frac{1}{{60}}} \right]\\ &= \frac{{11}}{2}\left[ {\frac{2}{{15}} + \frac{1}{6}} \right]\\& = \frac{{11}}{2}\left[ {\frac{{4 + 5}}{{30}}} \right]\\ &= \frac{{11}}{2} \times \frac{3}{{10}}\\& = \frac{{33}}{{20}}\end{align}

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