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.

Steps:

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,

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