# NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.3

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## Chapter 5 Ex.5.3 Question 1

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

### 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}

## Chapter 5 Ex.5.3 Question 2

Find the sums given below

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

(ii) $$34 + 32 + 30 + \dots+ 10$$

(iii) $$− 5 + (− 8) + (− 11) + \dots + (− 230)$$

### Solution

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

What is Known?

The AP $$7 + 10 \frac{1}{2} + 14 + \dots + 84$$

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) $$34 + 32 + 30 + \dots+ 10$$

What is Known?

The AP $$34 + 32 + 30 +\dots+ 10$$

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)

$$(−5) + (−8) + (−11) +\dots + (−230)$$

What is Known?

The AP

$$(−5) + (−8) + (−11) + \dots + (−230)$$

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}

## Chapter 5 Ex.5.3 Question 3

In an AP

(i) Given $$a = 5, d = 3, {a_n} = 50$$, find $$n$$ and $${S_n}$$.

(ii) Given $$a = 7, {a_{13}} = 35$$, find $$d$$ and $${S_{13}}$$.

(iii) Given $${a_{12}} = 37, d = 3$$, find $$a$$ and $${S_{12}}$$.

(iv) Given $${a_3} = 15, {S_{10}} = 125$$, find $$d$$ and $${a_{10}}$$.

(v) Given, find $$a$$ and $${a_9}$$.

(vi) Given $$a = 2, d = 8, {S_n} = 90$$, find $$n$$ and $${a_n}$$. $$d = 5, {S_9} = 75$$

(vii) Given $$a = 8, {a_n} = 62, {S_n} = 210$$, find $$n$$ and $$d$$.

(viii) Given $${a_n} = 4, d = 2, {S_n} = - 14$$, find $$n$$ and $$a$$.

(ix) Given $$a = 3, n = 8, S = 192$$, find $$d$$.

(x) Given $$l = 28, S = 144$$ and there are total $$9$$ terms. Find $$a$$.

### Solution

(i) Given $$a = 5, d = 3, {a_n} = 50$$, find $$n$$ and $${S_n}$$.

What is Known?

$$a = 5, d = 3, {a_n} = 50$$

What is Unknown?

$$n$$ and $${S_n}$$.

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, $$a = - 5$$
• Common difference, $$d = 3$$
• nth term, $$l = {a_n} = 50$$

As $$\,{a_n} = a + \left( {n - 1} \right)d$$

\begin{align}50&=5+\left( n-1 \right)3 \\45&=\left( n-1 \right)3 \\15&=n-1 \\n&=16 \\\end{align}

\begin{align}{S_n} &= \frac{n}{2}\left[ {a + l} \right]\\{S_{16}} &= \frac{{16}}{2}\left[ {5 + 50} \right]\\& = 8 \times 55\\ &= 440\end{align}

(ii) Given $$a = 7, {a_{13}} = 35$$, find $$d$$ and $${S_{13}}$$.

What is Known?

$$a = 7, {a_{13}} = 35$$,

What is Unknown?

$$d$$ and $${S_{13}}$$.

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, $$a = 7$$
• $$13^\rm{th}$$ term, $$l = {a_{13}} = 35$$

As $$\,{a_n} = a + \left( {n - 1} \right)d$$

\begin{align}{{a}_{13}}&=a+\left( 13-1 \right)d \\35&=7+12d \\35-7&=12d \\d&=\frac{28}{12} \\ d& =\frac{7}{3} \\\end{align}

\begin{align}{S_n} &= \frac{n}{2}\left[ {a + l} \right]\\{S_{13}} &= \frac{{13}}{2}\left[ {7 + 35} \right]\\ &= \frac{{13}}{2} \times 42\\ &= 13 \times 21\\ &= 273\end{align}

(iii) Given $${a_{12}} = 37, d = 3$$, find $$a$$ and $${S_{12}}$$.

What is Known?

$${a_{12}} = 37, d = 3$$

What is Unknown?

$$a$$ and$${S_{12}}.$$

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, is the common difference and $$n$$is the number of terms and $$l$$ is the last term.

Steps:

Given,

• $$12$$th term, $${a_{12}} = 37$$
• Common Difference, $$d = 3$$

As $$\,{a_n} = a + \left( {n - 1} \right)d$$

\begin{align}\,{a_{12}} &= a + \left( {12 - 1} \right)3\\37 &= a + 33\\a &= 4\end{align}

\begin{align}{S_n}& = \frac{n}{2}\left[ {a + l} \right]\\{S_{12}} &= \frac{{12}}{2}\left[ {4 + 37} \right]\\{S_{12}}& = 6 \times 41\\{S_{12}} &= 246\end{align}

(iv) Given $${a_3} = 15, {S_{10}} = 125$$, find $$d$$ and $${a_{10}}.$$

What is Known?

$${a_3} = 15, {S_{10}} = 125$$

What is Unknown?

$$d$$ and $${a_{10}}.$$

Reasoning:

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

Steps:

Given,

• 3rd term, $${a_3} = 15$$
• Sum up to ten terms, $${S_{10}} = 125$$

As $$\,{a_n} = a + \left( {n - 1} \right)d$$

\begin{align}{a_3} &= a + \left( {3 - 1} \right)d\\15 &= a + 2d \qquad \qquad \dots\text{(i)}\end{align}

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{10}} &= \frac{{10}}{2}\left[ {2a + \left( {10 - 1} \right)d} \right]\\125 &= 5\left[ {2a + 9d} \right]\\25 &= 2a + 9d \quad \dots\text{(ii)}\end{align}

On multiplying equation (i) by $$2,$$ we obtain

$30 = 2a + 4d \qquad \qquad \dots \text{(iii)}$

On subtracting equation (iii) from Equation (ii), we obtain

\begin{align} - 5 &= 5d\\d &= - 1\end{align}

From equation (i),

\begin{align}15 &= a + 2\left( { - 1} \right)\\15&= a-2\\a& = 17\end{align}

\begin{align}{a_{10}}& = a + \left( {10 - 1} \right)d\\{a_{10}}& = 17 + 9( - 1)\\{a_{10}} &= 17 - 9\\{a_{10}} &= 8\end{align}

(v) Given $$d = 5, {S_9} = 75$$, find $$a$$ and $${a_9}.$$

What is Known?

$$d = 5, {S_9} = 75$$

What is Unknown?

$$a$$ and $${a_9}.$$

Reasoning:

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

Steps:

Given,

• Common difference, $$d = 5$$
• Sum up to nine terms, $${S_9} = 75$$

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

\begin{align}{S_9} &= \frac{9}{2}\left[ {2a + \left( {9 - 1} \right)5} \right]\\75 &= \frac{9}{2}\left( {2a + 40} \right)\\25 &= 3\left( {a + 20} \right)\\25 &= 3a + 60\\a &= \frac{{ - 35}}{3}\end{align}

We know that $$n\rm{th}$$ term of the AP series, $${a_n} = a + \left( {n - 1} \right)d$$

\begin{align}{a_9} &= a + \left( {9 - 1} \right) \times 5\\ &= \frac{{ - 35}}{3} + 8 \times 5\\ &= \frac{{ - 35}}{3} + 40\\ &= \frac{{ - 35 + 120}}{3}\\ &= \frac{{85}}{3}\end{align}

(vi) Given $$a = 2, d = 8, {S_n} = 90$$, find $$n$$ and $${a_n}.$$

What is Known?

$$a = 2, d = 8, {S_n} = 90$$

What is Unknown?

$$n$$ and $${a_n}.$$

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, $$a = 2$$
• Common difference, $$d = 5$$
• Sum up to nth terms, $${S_n} = 90$$

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

\begin{align}90 = \frac{n}{2}\left[ {4 + \left( {n - 1} \right)8} \right]\\90{\rm{ }} = n{\rm{ }}[2 + (n - 1)4]\\90 = n\left[ {2 + 4n - 4} \right]\\90 = n\left[ {4n - 2} \right]\\90 = 4{n^2} - 2n\\4{n^2} - 2n - 90 = 0\\4{n^2} - 20n + 18n - 90 = 0\\4n\left( {n - 5} \right) + 18\left( {n - 5} \right) = 0\\\left( {n - 5} \right)\left( {4n + 18} \right) = 0\end{align}

Either $$(n - 5) = 0$$ or $$(4n + 18) = 0$$

$$n = 5\,$$ or $$n = \frac{{ - 9}}{2}$$

However, $$n$$ can neither be negative nor fractional.

Therefore, $$n = 5\,$$

\begin{align}{a_n} &= a + \left( {n - 1} \right)d\\{a_5} &= 2 + \left( {5 - 1} \right)8\\{a_5} &= 2 + 4 \times 8\\{a_5} &= 2 + 32\\{a_5} &= {\rm{ }}34\end{align}

(vii) Given $$a = 8, {a_n} = 62, {S_n} = 210$$, find and $$d$$.

What is Known?

$$a = 8, {a_n} = 62, {S_n} = 210$$

What is Unknown?

$$n$$ and $$d.$$

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, $$a = 8$$
• $$n^\rm{th}$$ term, $$l = {a_n} = 62$$
• Sum up to nth terms, $${S_n} = 210$$

As $${S_n} = \frac{n}{2}\left[ {a + l} \right]$$

\begin{align}210& = \frac{n}{2}\left[ {8 + 62} \right]\\210 &= \frac{n}{2} \times 70\\n &= 6\end{align}

We know that $$n^\rm{th}$$ term of the AP series, $${a_n} = a + \left( {n - 1} \right)d$$

\begin{align}62 &= 8 + \left( {6 - 1} \right)d\\62 - 8 &= 5d\\54 &= 5d\\d &= \frac{{54}}{5}\end{align}

(viii) Given $${a_n} = 4,{\rm{ }}d = 2,{\rm{ }}{S_n} = - {\rm{ }}14,$$ find $$n$$ and $$a$$.

What is Known?

$${a_n} = 4, d = 2, {S_n} = - 14$$

What is Unknown?

$$n$$ and $$a$$

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,

• Common difference, $$d = 2$$
• $$n^\rm{th}$$ term, $$l = {a_n} = 4$$
• Sum up to nth terms, $${S_n} = - 14$$

We know that nth term of AP series,$$\,{a_n} = a + \left( {n - 1} \right)d$$

\begin{align}4 &= a + (n - 1)2\\4 &= a + 2n - 2\\a &= 6 - 2n \qquad \dots(1)\end{align}

\begin{align}{S_n} &= \frac{n}{2}\left[ {a + l} \right]\\ - 14 &= \!\frac{n}{2}\left[ {6 \!- \!2n \!+\! 4} \right]\dots[\text{from(1)}]\\ - 14& = n\left( {5 - n} \right)\\ - 14 &= 5n - {n^2}\\{n^2} \!-\! 5n \!-\! 14\! &= 0\\\begin{bmatrix}{n^2}\! - \!7n + \\2n - 14 \end{bmatrix}&= 0\\\begin{bmatrix}\left( {n - 7} \right) + \\2\left( {n - 7} \right) \end{bmatrix}&= 0\\\left( \!{n \!-\! 7\!} \right)\left( \!{n \!+\! 2\!} \right) &= 0\end{align}

Either $$n - 7 = 0$$ or $$n + 2 = 0$$

$$n=7$$ or $$n = - 2$$

However, $$n$$ can neither be negative nor fractional.

Therefore, $$n=7$$

From equation (1), we obtain

\begin{align}a&= 6 - 2n\\a& = 6 - 2 \times 7\\a& = 6-14\\a& = - 8\end{align}

(ix) Given $$a = 3, n = 8, S = 192$$, find $$d$$.

What is Known?

$$a = 3, n = 8, S = 192$$

What is Unknown?

$$d$$

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, $$a = 3$$
• Number of terms, $$n = 8$$
• Sum up to nth terms, $${S_n} = 192$$

\begin{align}{S_n}& = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\192 &= \frac{8}{2}\left[ {2 \times 3 + \left( {8 - 1} \right)d} \right]\\192 &= 4\left[ {6 + 7d} \right]\\48 &= 6 + 7d\\42 &= 7d\\d &= 6\end{align}

(x) Given $$l = 28, S = 144$$ and there are total $$9$$ terms. Find $$a$$.

What is Known?

$$l = 28, S = 144, n = 9$$

What is Unknown?

$$a$$

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 $${S_n} = \frac{n}{2}\left[ {a + l} \right]$$

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,

• Last term, $$l = {a_n} = 28$$
• Number of terms, $$n = 9$$
• Sum up to nth terms, $${S_n} = 192$$

\begin{align}{S_n} &= \frac{n}{2}\left( {a + l} \right)\\144 &= \frac{9}{2}\left( {a + 28} \right)\\32 &= a + 28\\a &= 4\end{align}

## Chapter 5 Ex.5.3 Question 4

How many terms of the AP. $$9, 17, 25 \dots$$ must be taken to give a sum of $$636$$?

### Solution

What is Known?

The AP and sum.

What is Unknown?

Number of 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, $$a = 9$$

• Common difference, $$d = 17 - 9 = 8$$

• Sum up to nth terms, $${S_n} = 636$$

We know that sum of $$n$$ terms of AP

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\636 &= \frac{n}{2}\left[ {2 \!\times \!9 \!+\! \left( {n - 1} \right)8} \right]\\636 &= \frac{n}{2}\left[ {18 + 8n - 8} \right]\\636& = \frac{n}{2}\left[ {10 + 8n} \right]\\636& = n\left[ {5 + 4n} \right]\\636& = 5n + 4{n^2}\\4{n^2} + 5n - 636 &= 0\\\begin{bmatrix} 4{n^2} + 53n - \\48n - 636\end{bmatrix} &= 0\\\begin{bmatrix}n\left( {4n + 53} \right) -\\ 12\left( {4n + 53} \right)\end{bmatrix} &= 0\4n + 53)(n - 12) &= 0\end{align} Either \(4n + 53 = 0 or $$n - 12 = 0$$

$$n = - \frac{{53}}{4}$$ or $$n = 12$$

$$n$$ cannot be $$\frac{{ - 53}}{4}$$. As the number of terms can neither be negative nor fractional, therefore, $$n = 12$$

## Chapter 5 Ex.5.3 Question 5

The first term of an AP is $$5,$$ the last term is $$45$$ and the sum is $$400.$$ Find the number of terms and the common difference.

### Solution

What is Known?

$$a,{\rm{ }}l$$, and $${S_n}$$

What is Unknown?

$$n$$ and $$l$$

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, $$a = 5$$
• Last term, $$l = 45$$
• Sum up to $$n\rm{th}$$ terms, $${S_n} = 400$$

We know that sum of $$n$$ terms of AP

\begin{align}{S_n} &= \frac{n}{2}\left( {a + l} \right)\\400& = \frac{n}{2}\left( {5 + 45} \right)\\400&= \frac{n}{2} \times 50\\n &= 16\end{align}

\begin{align}l &= {a_n} = a + \left( {n - 1} \right)d\\45 &= 5 + \left( {16 - 1} \right)d\\40 &= 15d\\d &= \frac{{40}}{{15}}\\d& = \frac{8}{3}\end{align}

## Chapter 5 Ex.5.3 Question 6

The first and the last term of an AP are $$17$$ and $$350$$ respectively. If the common difference is $$9,$$ how many terms are there and what is their sum?

### Solution

What is known?

$$a,{\rm{ }}l$$, and $$d$$

What is unknown?

$$n$$ and $${S_n}$$

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, $$a = 17$$
• Last term, $$l = 350$$
• Common difference, $$d = 9$$

We know that $$n\rm{th}$$ term of AP,  $$\,l = {a_n} = a + \left( {n - 1} \right)d$$

\begin{align}350& = 17 + \left( {n - 1} \right)9\\333 &= \left( {n - 1} \right)9\\\left( {n - 1} \right) &= 37\\n &= 38\end{align}

Sum of $$n$$ terms of AP,

\begin{align}{S_n} &= \frac{n}{2}\left( {a + l} \right)\\{S_{38}} &= \frac{{38}}{2}\left( {17 + 350} \right)\\ \, &= 19 \times 367\\ &= 6973\end{align}

Thus, this A.P. contains $$38$$ terms and the sum of the terms of this A.P. is $$6973.$$

## Chapter 5 Ex.5.3 Question 7

Find the sum of first $$22$$ terms of an AP in which $$d = 7$$ and $$22\rm{nd}$$ term is $$149.$$

### Solution

What is Known?

$$d$$ and $${a_{22}}$$

What is Unknown?

$${S_{22}}$$

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,

22nd term, $$l = {a_{22}} = 149$$

Common difference, $$d = 7$$

We know that $$n^\rm{th}$$ term of AP, $${a_n} = a + \left( {n - 1} \right)d$$

\begin{align}{{a}_{22}}&= a+\left( 22-1 \right)d \\149&=a+21\times 7 \\ 149&=a+147 \\ a &=2 \\\end{align}

\begin{align}{S_n}&= \frac{n}{2}\left( {a + l} \right)\\ &= \frac{{22}}{2}\left( {2 + 149} \right)\\& = 11 \times 151\\ &= 1661\end{align}

## Chapter 5 Ex.5.3 Question 8

Find the sum of first $$51$$ terms of an AP whose second and third terms are $$14$$ and $$18$$ respectively.

### Solution

What is known?

$${a_2}$$ and $${a_3}$$

What is unknown?

$${S_{51}}$$

Reasoning:

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

2nd term, $${a_2} = 14$$

3rd term, $${a_3} = 18$$

Common difference, $$d = {a_3} - {a_2} = 18 - 14 = 4$$

We know that $$n\rm{th}$$ term of AP, $${a_n} = a + \left( {n - 1} \right)d$$

\begin{align}{a_2} &= a + d\\14 &= a + 4\\a& = 10\end{align}

Sum of $$n$$ terms of AP series,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{51}} &= \frac{{51}}{2}\left[ {2 \times 10 + \left( {51 - 1} \right)4} \right]\\ &= \frac{{51}}{2}\left[ {20 + 50 \times 4} \right]\\ &= \frac{{51}}{2} \times 220\\ &= 51 \times 110\\ &= 5610\end{align}

## Chapter 5 Ex.5.3 Question 9

If the sum of first $$7$$ terms of an AP is $$49$$ and that of $$17$$ terms is $$289,$$ find the sum of first $$n$$ terms.

### Solution

What is Known?

$${S_7}$$ and $${S_{17}}$$

What is Unknown?

$${S_n}$$

Reasoning:

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

Sum of first 7 terms, $${S_7} = 49$$

Sum of first 17 terms, $${S_{17}} = 289$$

We know that sum of $$n$$ term of AP is,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_7} &= \frac{7}{2}\left[ {2a + \left( {7 - 1} \right)d} \right]\\49 &= \frac{7}{2}\left[ {2a + 6d} \right]\\a + 3d &= 7\qquad \qquad ....\left( \rm{i} \right)\\\\{S_{17}}&= \frac{{17}}{2}\left[ {2a + \left( {17 - 1} \right)d} \right]\\289 &= \frac{{17}}{2}\left[ {2a + 16d} \right]\\a + 8d &= 17\qquad \qquad ....\left( {\rm{ii}} \right)\end{align}

Subtracting equation (i) from equation (ii),

\begin{align}5d &= 10\\d &= 2\end{align}

From equation (i),

\begin{align}7& = a + 3 \times 2\\7 &= a + 6\\a&= 1\end{align}

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\ &= \frac{n}{2}\left[ {2 \times 1 + \left( {n - 1} \right)2} \right]\\ &= \frac{n}{2}\left[ {2 + 2n - 2} \right]\\ &= \frac{n}{2} \times 2n\\& = {n^2}\end{align}

## Chapter 5 Ex.5.3 Question 10

Show that $${a_1},{\rm{ }}{a_2},...{\rm{ }},{\rm{ }}{a_n},{\rm{ }}...$$form an AP where $${a_n}$$ is defined as below

(i) $${a_n} = 3 + 4n$$

(ii) $${a_n} = 9 - 5n$$

Also find the sum of the first $$15$$ terms in each case.

### Solution

(i) $${a_n} = 3 + 4n$$

What is known?

$$\,{a_n} = 3 + 4n$$

What is unknown?

Whether $${a_1},{\rm{ }}{a_2},...{\rm{ }},{\rm{ }}{a_n},{\rm{ }}...$$ form an AP

Reasoning:

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

Steps:

Given,

$$nth$$ term, $${a_n} = 3 + 4n$$

\begin{align}{a_1} &= 3 + 4 \times 1\\&= 7\\{a_2} &= 3 + 4 \times 2 = 3 + 8\\&= 11\\{a_3} &= 3 + 4 \times 3 = 3 + 12\\&= 15\\{a_4} &= 3 + 4 \times 4 = 3 + 16\\&= 19\end{align}

It can be observed that

\begin{align}{a_2} - {a_1} &= 11 - 7 \\&= 4\\{a_3} - {a_2} &= 15 - 11\\& = {\rm{ }}4\\{a_4} - {a_3} &= 19 - 15\\& = {\rm{ }}4\end{align}

i.e., the difference of $${a_n}$$ and $${a_{n - 1}}$$ is constant.

Therefore, this is an AP with common difference as $$4$$ and first term as $$7.$$

Sum of $$n$$ terms,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{15}} &= \frac{{15}}{2}\left[ {2 \times 7 + \left( {15 - 1} \right)4} \right]\\ &= \frac{{15}}{2}\left[ {14 + 14 \times 4} \right]\\ &= \frac{{15}}{2} \times 70\\ &= 15 \times 35\\ &= 525\end{align}

(ii) $${a_n} = 9 - 5n$$

What is known?

$${a_n} = 9 - 5n$$

What is unknown?

Whether $${a_1},{\rm{ }}{a_2},...{\rm{ }},{\rm{ }}{a_n},{\rm{ }}...$$ form an AP

Reasoning:

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

Steps:

Given,

$$nth$$ term, $${a_n} = 9 - 5n$$

\begin{align}{a_1} &= 9 - 5 \times 1\\&= 9 - 5 \\&= 4\\{a_2} &= 9 - 5 \times 2\\&= 9 - 10\\&= - 1\\{a_3} &= 9 - 5 \times 3 \\&= 9 - 15\\&= - 6\\{a_4}&= 9 - 5 \times 4\\&= 9 - 20\\&= - 11\end{align}

It can be observed that

\begin{align}{a_2} - {a_1} &= ( - 1{\rm{ )}} - {\rm{ }}4 \\&= - 5\\{a_3} - {a_2}& = ( - 6) - ( - 1)\\& = - 5\\{a_4} - {a_3} &= ( - 11) - ( - 6) \\&= - 5\end{align}

i.e., the difference of $${a_n}$$ and $${a_{n - 1}}$$ is constant.

Therefore, this is an A.P. with common difference as $$(−5)$$ and first term as $$4.$$

Sum of n terms of AP,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{15}} &= \frac{{15}}{2}\left[ {2 \times 4 + \left( {15 - 1} \right)\left( { - 5} \right)} \right]\\ &= \frac{{15}}{2}\left[ {8 + 14\left( { - 5} \right)} \right]\\ &= \frac{{15}}{2}\left[ {8 - 70} \right]\\ &= \frac{{15}}{2}\left[ { - 62} \right]\\& = - 465\end{align}

## Chapter 5 Ex.5.3 Question 11

If the sum of the first $$n$$ terms of an AP is $$4n - {n^2}$$, what is the first term (that is $$S_1$$)?

What is the sum of first two terms?

What is the second term? Similarly find the $$3\rm{rd,}$$ the$$10\rm{th}$$ and the $$n\rm{th}$$ terms.

### Solution

What is Known?

$${S_n} = 4{n^2} - {n^2}$$

What is Unknown?

$${S_1},{\rm{ }}{S_2},{\rm{ }}{a_2},{\rm{ }}{a_3},{\rm{ }}{a_{10}}$$ and $${a_n}$$

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,

Sum of first $$n$$ terms, $${S_n} = 4n - {n^2}$$

Therefore,

Sum of first term,

$a = {S_1} = 4 \times 1 - {1^2} = 4 - 1 = 3$

Sum of first two terms,

${S_2} = 4 \times 2 - {2^2} = 8 - 4 = 4$

Sum of first three terms,

${S_3} = 4 \times 3 - {3^2} = 12 - 9 = 3$

Second term,

${a_2} = {S_2} - {S_1} = 4 - 3 = 1$

Third term,

${a_3} = {S_3} - {S_2} = 3 - 4 = - 1$

Tenth term,

${a_{10}} = {S_{10}} - {S_9}$

\begin{align}& = \left( {4 \times 10 - {{10}^2}} \right) - \left( {4 \times 9 - {9^2}} \right)\\ &= \left( {40 - 100} \right) - \left( {36 - 81} \right)\\& = - 60 + 45\\ &= - 15\end{align}

$$n\rm{th}$$ term, $${a_n} = {S_n} - {S_{n - 1}}$$

\begin{align}&\!=\!\left[ {4n - {n^2}} \right]\!-\!\left[\!{4\left( {n\!-\!1} \right)\!-\!{{\left( {n\!-\!1} \right)}^2}} \!\right]\\ &= 4n - {n^2} - 4n + 4 + {\left( {n - 1} \right)^2}\\ &= 4 - {n^2} + {n^2} - 2n + 1\\ &= 5 - 2n\end{align}

Hence, the sum of first two terms is $$4.$$

The second term is $$1.$$

$$3\rm{rd},\; 10\rm{th},$$ and $$n\rm{th}$$ terms are $$−1, −15,$$ and $$(5 - 2n)$$ respectively.

## Chapter 5 Ex.5.3 Question 12

Find the sum of first $$40$$ positive integers divisible by $$6.$$

### Solution

What is Known?

Positive integers divisible by $$6$$

What is Unknown?

Sum of first $$40$$ positive integers divisible by $$6,$$ $${S_{40}}$$

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:

The positive integers that are divisible by $$6$$ are $$6, 12, 18, 24, \dots$$

It can be observed that these are making an AP

Hence,

• First term, $$a = 6$$
• Common difference, $$d = 6$$
• Number of terms, $$n = 40$$

As we know that Sum of $$n$$ terms,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{40}} &= \frac{{40}}{2}\left[ {2 \times 6 + \left( {40 - 1} \right)6} \right]\\ &= 20\left[ {12 + 39 \times 6} \right]\\ &= 20\left[ {12 + 234} \right]\\& = 20 \times 246\\ &= 4920\end{align}

## Chapter 5 Ex.5.3 Question 13

Find the sum of first $$15$$ multiples of $$8.$$

### Solution

What is Known?

Multiples of $$8$$

What is Unknown?

Sum of first $$15$$ multiples of $$8,$$ $${S_{15}}$$

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:

The multiples of $$8$$ are $$8, 16, 24, 32, \dots$$

These are in an A.P.,

Hence,

• First term, $$a = 8$$
• Common difference, $$d = 8$$
• Number of terms, $$n = 15$$

As we know that Sum of $$n$$ terms,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{15}} &= \frac{{15}}{2}\left[ {2 \times 8 + \left( {15 - 1} \right)8} \right]\\ &= \frac{{15}}{2}\left[ {16 + 14 \times 8} \right]\\ &= \frac{{15}}{2}\left[ {16 + 112} \right]\\& = \frac{{15}}{2} \times 128\\ &= 15 \times 64\\& = 960\end{align}

## Chapter 5 Ex.5.3 Question 14

Find the sum of the odd numbers between $$0$$ and $$50.$$

### Solution

What is Known?

Odd numbers between $$0$$ and $$50$$

What is Unknown?

Sum of the odd numbers between $$0$$ and $$50$$

Reasoning:

Sum of the first

$$n$$ terms of an AP is given by $${S_n} = \frac{n}{2}\left[ {a + l} \right]$$, 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:

The odd numbers between $$0$$ and $$50$$ are $$1, \,3,\, 5,\, 7,\, 9 \,... \,49$$

Therefore, it can be observed that these odd numbers are in an A.P.

Hence,

• First term, $$a = 1$$
• Common difference, $$d = 2$$
• Last term, $$l = 1$$

We know that nth term of AP, $$\,{a_n} = l = a + \left( {n - 1} \right)d$$

\begin{align}49 &= 1 + \left( {n - 1} \right)2\\48& = 2\left( {n - 1} \right)\\n - 1 &= 24\\n &= 25 \end{align}

We know that sum of $$n$$ terms of AP,

\begin{align}{S_n} &= \frac{n}{2}\left( {a + l} \right)\\{S_{25}} &= \frac{{25}}{2}\left( {1 + 49} \right)\\ &= \frac{{25}}{2} \times 50\\& = 25 \times 25\\ &= 625\end{align}

## Chapter 5 Ex.5.3 Question 15

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: $$\rm{Rs.}\, 200$$ for the first day, $$\rm{Rs.}\, 250$$ for the second day, $$\rm{Rs.}\,300$$ for the third day, etc., the penalty for each succeeding day being $$\rm{Rs.}\,50$$ more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by $$30$$ days.

### Solution

What is Known?

The penalty for delay of completion by a day, $$\rm{Rs} \,200$$ and $$\rm{Rs} \,50$$ more each succeeding day.

What is Unknown?

Amount has to pay as a penalty.

Reasoning:

General form of an arithmetic progression is $$a,{\rm{ }}\left( {a + d} \right),{\rm{ }}\left( {a + 2d} \right),{\rm{ }}\left( {a + 3d} \right).$$

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:

Penalty for $$1st$$ day $$\rm{Rs}.\, 200.$$

Penalty for $$2nd$$ day $$\rm{Rs}.\,250$$

Penalty for $$3rd$$ day $$\rm{Rs}.\,300$$

By observation that these penalties are in an A.P. having first term as $$200$$ and common difference as $$50$$ and number of terms as $$30.$$

\begin{align}a &= 200\\d &= 50\\n &= 30\end{align}

Penalty that has to be paid if he has delayed the work by $$30$$ days

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{30}} &= \frac{{30}}{2}\left[ {2 \times 200 + \left( {30 - 1} \right)50} \right]\\ &= 15\left[ {400 + 1450} \right]\\ &= 15 \times 1850\\& = 27750\end{align}

Therefore, the contractor has to pay $$\rm{Rs}. 27750$$ as penalty.

## Chapter 5 Ex.5.3 Question 16

A sum of $$\rm{Rs}\, 700$$ is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is $$\rm{Rs}\,20$$ less than its preceding prize, find the value of each of the prizes.

### Solution

What is Known?

$$7$$ cash prizes are given, and each prize is $$\rm{Rs}\,20$$ less than its preceding prize.

What is Unknown?

Value of each of the prizes

Reasoning:

General form of an arithmetic progression is $$a,\left( {a + d} \right),\left( {a + 2d} \right),\left( {a + 3d} \right),\dots$$

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:

Let the cost of 1st prize be $$x$$.

Then the cost of 2nd prize $$=x-20$$

And the cost of 3rd prize $$= x - 40$$

Prizes are $$x,{\rm{ }}\left( {x - 20} \right),{\rm{ }}\left( {x - 40} \right),\dots$$

By observation that the costs of these prizes are in an A.P., having common difference as $$−20$$ and first term as $$x$$.

\begin{align} a&=x \\ d& =-20 \\\end{align}

Given that, $${S_7} = 700$$

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

\begin{align}\frac{7}{2}\left[ {2x + \left( {7 - 1} \right)d} \right] &= 700\\\left[ {2x + \left( 6 \right) \times \left( { - 20} \right)} \right] &= 200\\x + 3 \times \left( { - 20} \right) &= 100\\x - 60 &= 100\\x &= 160\end{align}

Therefore, the value of each of the prizes was $$\rm{Rs}\,160,$$ $$\rm{Rs} \,140,$$ $$\rm{Rs}\, 120, \;\rm{Rs}\,100, \;\rm{Rs}\,80, \;\rm{Rs}\,60,$$ and $$\rm{Rs} \,40.$$

## Chapter 5 Ex.5.3 Question 17

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant $$1$$ tree, a section of class II will plant $$2$$ trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

### Solution

What is Known?

The number of trees planted by $$3$$ sections of each class (I to XII).

What is Unknown?

Number of trees planted by the students.

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:

 Class I Class II Class III …………………… Class XII Section $$A$$ $$1$$ $$2$$ $$3$$ …………………… $$12$$ Section $$B$$ $$1$$ $$2$$ $$3$$ …………………… $$12$$ Section $$C$$ $$1$$ $$2$$ $$3$$ …………………… $$12$$ Total $$3$$ $$6$$ $$9$$ $$36$$

It can be observed that the number of trees planted by the students are in an AP.

$3, 6, 9, 12, 15,\dots\dots36$

• First term $$a = 3$$
• Common difference, $$d = 6 - 3 = 3$$
• Number of terms, $$n = 12$$

We know that sum of $$n$$ terms of AP,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{12}} &= \frac{{12}}{2}\left[ {2 \times 3 + \left( {12 - 1} \right) \times 3} \right]\\& = 6\left[ {6 + 11 \times 3} \right]\\ &= 6\left[ {6 + 33} \right]\\ &= 6 \times 39\\ &= 234\end{align}

Therefore, $$234$$ trees will be planted by the students.

## Chapter 5 Ex.5.3 Question 18

A spiral is made up of successive semicircles, with centres alternately at $$A$$ and $$B,$$ starting with centre at $$A$$ of radii $$0.5,$$ $$1.0 \,\rm{cm}, 1.5 \,\rm{cm}, 2.0 \,\rm{cm}, .........$$as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?

\begin{align}\left[ {{\rm{Take}}\,\,\pi = \frac{{22}}{7}} \right] \end{align} ### Solution

What is Known?

Radii of the $$13$$ semicircles.

What is Unknown?

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:

Semi-perimeter of circle, $$l = \pi r$$

\begin{align}{{l}_{1}}& =\pi \times \left( 0.5\,\,\text{cm} \right)=0.5\pi \,\,\text{cm } \\ {{l}_{2}}&=\pi \times \left( 1\,\,\text{cm} \right)=\pi \,\,\text{cm} \\ {{l}_{3}} &=\pi \times \left( 1.5\,\,\text{cm} \right)=1.5\pi \,\text{cm} \\\end{align}

Therefore, $${l_1},{\rm{ }}{l_2},{\rm{ }}{l_3},$$ i.e. the lengths of the semi-circles are in an A.P.,

\begin{align}&0.5\pi ,\pi ,1.5\pi ,2\pi\dots\dots\dots\\&a = 0.5\pi \\&d = \pi - 0.5\pi = 0.5\pi\end{align}

We know that the sum of $$n$$ terms of an A.P. is given by

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{13}} &\!=\!\frac{{13}}{2}\left[ {2 \!\times\! \left( {0.5\pi } \right)\!+\!\left( {13\!-\!1} \right)\left( {0.5\pi } \right)}\!\right]\\ &= \frac{{13}}{2}\left[ {\pi + 6\pi } \right]\\ &= \frac{{13}}{2} \times 7\pi \\ &= \frac{{13}}{2} \times 7 \times \frac{{22}}{7}\\ &= 143\end{align}

Therefore, the length of such spiral of thirteen consecutive semi-circles will be $$143\, \rm{cm.}$$

## Chapter 5 Ex.5.3 Question 19

$$200$$ logs are stacked in the following manner: $$20$$ logs in the bottom row, $$19$$ in the next row, $$18$$ in the row next to it and so on. In how many rows are the $$200$$ logs placed and how many logs are in the top row?

### Solution

What is Known?

Stack of $$200$$ logs, $$20$$ logs in the bottom row, $$19$$ in the next row, $$18$$ in the row next to it and so on.

What is Unknown?

Number of rows and number of logs in the top row

Reasoning:

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

Steps:

It can be observed that the numbers of logs in rows are in an A.P.

$20, 19, 18, ...$

For this A.P.,

• First terms, $$a = 20$$
• Common difference, $$d = 19 - 20 = - 1$$
• Sum of the n terms, $${S_n} = 200$$

We know that sum of $$n$$ terms of AP,

\begin{align}{{S}_{n}}&\!=\!\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\200&\!=\!\frac{n}{2}\left[ \begin{array} & 2\times 20+ \\ \left( n-1 \right)\left( -1 \right) \end{array} \right] \\ 400&\!=\!n\left[ 40-n+1 \right] \\ 400&\!=\!n\left[ 41-n \right] \\ 400&\!=\!41n-{{n}^{2}} \\ {{n}^{2}}-41n+400&\!=\!0 \\ \left[ \begin{array} & {{n}^{2}}-16n- \\ 25n+400 \\ \end{array} \right]&\!=\!0 \\ \left[ \begin{array} & n\left( n-16 \right)- \\ 25\left( n-16 \right) \\ \end{array} \right]&\!=\!0 \\ \left( n-16 \right)\left( n-25 \right)&\!=\!0 \\ \end{align}

Either $$\left( {n - 16} \right) = 0$$ or $$\left( {n - 25} \right) = 0$$

$$\therefore n = 16$$ or $$n = 25$$

\begin{align}{a_n} &= a + \left( {n - 1} \right)d\\{a_{16}}&= 20 + \left( {16 - 1} \right) \times \left( { - 1} \right)\\{a_{16}}& = 20 - 15\\{a_{16}} &= {\rm{ }}5\end{align}

Similarly,

\begin{align}{a_{25}}&= 20 + \left( {25 - 1} \right) \times \left( { - 1} \right)\\{a_{25}} &= 20 - 24\\{a_{25}} &= - 4\end{align}

Clearly, the number of logs in $$16$$th row is $$5.$$ However, the number of logs in $$25\rm{th}$$ row is negative $$4,$$ which is not possible.

Therefore, $$200$$ logs can be placed in $$16$$ rows.

The number of logs in the top $$(16^\rm{th})$$ row is $$5.$$

## Chapter 5 Ex.5.3 Question 20

In a potato race, a bucket is placed at the starting point, which is $$5 \,\rm{m}$$ from the first potato and other potatoes are placed $$3 \,\rm{m}$$ apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is $$2 × 5 + 2 × (5 + 3)$$]

### Solution

What is Known?

Distance of $$10$$ potatoes from the bucket, first potato is $$5 \,\rm{m}$$ away from bucket and other potatoes are placed $$3\,\rm{m}$$ apart in a straight line.

What is Unknown?

Total distance the competitor has to run.

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:

The distances of potatoes are as follows.

$5, 8, 11, 14, ...$

It can be observed that these distances are in A.P.

• First term, $$a = 5$$
• Common difference, $$d = 8 - 5 = 3$$
• Number of terms, $$n = 10$$

We know that the sum of $$n$$ terms,

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{10}} &= \frac{{10}}{2}\left[ {2 \times 5 + (10 - 1) \times 3} \right]\\{S_{10}} &= 5\left[ {10 + 27} \right]\\{S_{10}} &= 5 \times 37\\{S_{10}} &= 185\end{align}

As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it.

Therefore, total distance that the competitor will run

$$= 2 \times 185\, \rm{m }= 370\,{\rm{m }}$$

Alternatively,

The distances of potatoes from the bucket are $$5, 8, 11, 14...$$

Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.

Therefore, distances to be run are

First potato, $$a = 2 \times 5\,\rm{m} = 10\rm{m}$$

Second potato,

${a_2} = 2 \times \left( {5 + 3} \right)\,\rm{m} = 16\,\rm{m}$

Third potato,

${a_3} = 2 \times \left( {5 + 2 \times 3} \right)\,\rm{m} = 22\,\rm{m}$

Number of potatoes, $$n = 10$$

$$10,\,16,\,22,\,28,\,34,\ldots \ldots \ldots$$

\begin{align}a &= 10\\d& = 16 - 10 = 6\\{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{10}} &= \frac{{10}}{2}\left[ {2 \times 10 + (10 - 1)6} \right]\\&= 5\left[ {20 + 54} \right]\\&= 5 \times 74\\&= 370\end{align}

Therefore, the competitor will run a total distance of $$370 \,\rm{m}.$$