# 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}