Ex.5.3 Q2 Arithmetic progressions Solutions - NCERT Maths Class 10

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Question

Find the sums given below

(i) \(7 + 10\frac{1}{2} + 14 + ............ + 84\)

(ii) \(34 + 32 + 30 + ........... + 10\)

(iii) \(− 5 + (− 8) + (− 11) + ............ + (− 230)\)

 Video Solution
Arithmetic Progressions
Ex 5.3 | Question 2

Text Solution

(i) \(7 +10\frac{1}{2}+ 14 + ............+ 84\)

What is Known?

The AP \(7 + 10 \frac{1}{2} + 14 + ............ + 84\)

What is Unknown?

Sum of the AP

Reasoning:

Sum of the first \(n\) terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) or \({S_n} = \frac{n}{2}\left[ {a + l} \right]\), and \(n\,\rm{th}\) term of an AP is\(\,{a_n} = a + \left( {n - 1} \right)d\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms and \(l\) is the last term.

Steps:

Given,

  • First term, \(a = 7\)
  • Common Difference, \(\begin{align} d = 10\frac{1}{2} - 7 = \frac{{21}}{2} - 7 = \frac{7}{2} \end{align}\)
  • Last term, \(l = 84\)

\[\begin{align}l &= {a_n} = a + \left( {n - 1} \right)d\\84 &= 7 + \left( {n - 1} \right)\frac{7}{2}\\77 &= (n - 1)\frac{7}{2}\\22 &= n - 1\\n &= 23\end{align}\]

Hence, \(n = 23\)

\[\begin{align}{S_n} &= \frac{n}{2}(a + l)\\{S_{23}} &= \frac{{23}}{2}\left[ {7 + 84} \right]\\ &= \frac{{23}}{2} \times 91\\ &= \frac{{2093}}{2}\\ &= 1046\frac{1}{2}\end{align}\]

(ii) \(34 + 32 + 30 + ........... + 10\)

What is Known?

The AP \(34 + 32 + 30 + ........... + 10\)

What is Unknown?

Sum of the AP

Reasoning:

Sum of the first \(n\) terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) or \({S_n} = \frac{n}{2}\left[ {a + l} \right]\), and \(nth\) term of an AP is \(\,{a_n} = a + \left( {n - 1} \right)d\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms and \(l\) is the last term.

Steps:

Given,

  • First term, \(a = 34\)
  • Common Difference, \(d = 32 - 34 = - 2\)
  • Last term, \(l = 10\)

\[\begin{align}l &= a + (n - 1)d\\10 &= 34 + (n - 1)( - 2)\\n - 1 &= 12\\n &= 13\end{align}\]

Hence, \(n = 13\)

\[\begin{align}{S_n} &= \frac{n}{2}(a + l)\\{S_{23}} &= \frac{{13}}{2}\left[ {34 + 10 } \right]\\
& = \frac{{13}}{2} \times 44\\ &= 13 \times 22\\&= 286\end{align}\]

(iii)  \((−5) + (−8) + (−11) + ............ + (−230)\)

What is Known?

The AP \((−5) + (−8) + (−11) + ............ + (−230)\)

What is Unknown?

Sum of the AP

Reasoning:

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

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms and \(l\) is the last term.

Steps:

Given,

  • First term, \(a = - 5\)
  • Common Difference, \(d = ( - 8) - ( - 5) = - 8 + 5 = - 3\)
  • Last term, \(l = - 230\)

\[\begin{align}l &= a + (n - 1)d\\ - 230 &= ( - 5) + (n - 1)( - 3)\\n - 1 &= \frac{{225}}{3}\\n &= 75 + 1\\n &= 76\end{align}\]

Hence, \(n = 76\)

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {a + l} \right]\\ &= \frac{{76}}{2}\left[ {\left( { - 5} \right) + \left( { - 230} \right)} \right]\\& = 38\left[ { - 235} \right]\\ &= - 8930\end{align}\]