# NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.2

## Chapter 5 Ex.5.2 Question 1

Fill in the blanks in the following table, given that $$a$$ is the first term and $$d$$ is the Common difference and \begin{align}{a_n}\end{align} the $$n^{th}$$ term of the AP.

 $$a$$ $$d$$ $$n$$ \begin{align}{a_n}\end{align} $$7$$ $$3$$ $$8$$ $$……..$$ $$- 18$$ $$……..$$ $$10$$ $$0$$ $$…….$$ $$-3$$ $$18$$ $$-5$$ $$-18. 9$$ $$2.5$$ $$……..$$ $$3.6$$ $$3.5$$ $$0$$ $$105$$ $$……..$$

### Solution

i) Reasoning:

\begin{align}{a_n} = a + \left( {n - 1} \right)d\end{align}

What is Known?

$$a = 7,d = 3,n = 8$$

What is Unknown?

\begin{align}{a_n}\end{align}

Steps:

\begin{align}{a_n} &= a + (n - 1)d\\ {a_8} &= 7 + (8 - 1)3\\& = 7 + 7 \times 3\\& = 7 + 21\\& = 28 \end{align}

The Answer is $$a_n = 28$$

ii) What is Known?

$$a = - 18,\,{a_n} = 0,n = 10$$

What is Unknown?

$$d.$$

Steps:

\begin{align}a_n &= a + (n - 1)d\\ 0 &= - 18 + (10 - 1)d\\ 0 &= - 18 + 9d\\ 9d &= 18\\ d &= 2 \end{align}

The Answer is \begin{align}d = 2\end{align}

iii) What is Known?

$$d = -3, \,a_n = -5 , n = 18$$

What is Unknown?

$$a$$

Steps:

\begin{align} {a_n} &= a + (n - 1)d\\ - 5 &= a + (18 - 1)( - 3)\\ - 5 &= a + 17 \times ( - 3)\\ - 5 &= a - 51\\ a &= 51 - 5\\ a &= 46 \end{align}

The Answer is $$a = 46$$

iv) What is Known?

$${a_n} = 3.6\,,d = 2.5\,,a = - 18.9$$

What is Unknown?

\begin{align}n. \end{align}

Steps:

\begin{align} {a_n}& = a + (n - 1)d\\ 3.6 &= - 18.9 + (n - 1)2.5\\ 3.6 + 18.9 &= 2.5(n - 1)\\ 22.5& = 2.5(n - 1)\\ n - 1 &= \frac{{22.5}}{{2.5}}\\ n - 1 &= \frac{{225}}{{25}}\\ n - 1 &= 9\\ n& = 10 \end{align}

The Answer is $$n = 10$$

v) What is Known?

$$a = 3.5,d = 0,n = 105$$

What is Unknown?

\begin{align}{a_n}\end{align}

Steps:

\begin{align} {a_n} &= a + (n - 1)d\\ {a_n} &= 3.5 + (105 - 1)(0)\\ {a_n} &= 3.5 + 104 \times 0\\ {a_n} &= 3.5 \end{align}

The Answer is $$a_n = 3.5$$

## Chapter 5 Ex.5.2 Question 2

Choose the correct choice in the following and justify:

i) $$30^{th}$$ term of the \begin{align}\text{AP: 10,7,4……. is}\end{align}

a) $$97$$

b) $$77$$

c) $$-77$$

d) $$-87$$

ii) $$11^{th}$$ term of \begin{align}\text{AP: }-3, - \frac{1}{2},2\, \rm{is:}\end{align}

a) $$28$$

b) $$22$$

c) $$– 38$$

d) $$- 48$$

### Solution

Reasoning:

$$n^{th}$$ term of a AP is \begin{align}\,{a_n} = a + \left( {n - 1} \right)d\end{align}

i)  What is  Known?

The AP.

What is Unknown?

$$30^{th}$$ term of the AP.

Steps:

The AP is $$10,7,4, …..$$

\begin{align}a = 10\end{align}

Common difference

\begin{align} &=a_{2}-a_{1} \\ &=7-10 \\ &=-3 \end{align}

\begin{align}{a_n} &= a + (n - 1)d\\{a_{30}} &= 10 + (30 - 1)( - 3)\\ &= 10 + (29)( - 3)\\ &= 10 - 87\\ &= - 77\end{align}

The correct option is \begin{align}c = -77\end{align}

$$30^{th}$$ term is \begin{align}-77.\end{align}

ii)  What is  Known?

The AP.

What is Unknown?

$$11^{th}$$ term of the AP.

Steps:

The AP is -3,\begin{align} - \frac{1}{2}\end{align} , $$2$$

Common difference

\begin{align}d &=a_{2}-a_{1} \\ &=-\frac{1}{2}-(-3) \\ &=-\frac{1}{2}+3 \\ &=\frac{-1+6}{2} \\ d &=\frac{5}{2} \end{align}

\begin{align}{a_{11}} &= - 3 + (11 - 1)\frac{5}{2}\\ &= - 3 + 10 \times \frac{5}{2}\\ &= - 3 + 25\\{a_{11}} &= 22\end{align}

The correct option is B. $$11^{th}$$ term is $$22.$$

## Chapter 5 Ex.5.2 Question 3

In the following AP, find the missing terms in the boxes:

i) \begin{align}2, \boxed{\;\;}, 26 \end{align}

ii) \begin{align}\boxed{\;\;}13, \boxed{\;\;}, 3\end{align}

iii) \begin{align}5, \boxed{\;\;}, \boxed{\;\;}, 9 \frac{1}{2}\end{align}

iv) \begin{align}-4, \boxed{\;\;}, \boxed{\;\;}, \boxed{\;\;}, \boxed{\;\;} \end{align}

v) \begin{align}\boxed{\;\;}, 38, \boxed{\;\;}, \boxed{\;\;}, \boxed{\;\;},-22 \end{align}

### Solution

Reasoning:

\begin{align}{a_n} = a + \left( {n - 1} \right)d\end{align}

i) What is Known?

The AP with missing terms.

What is Unknown?

Second term.

Steps:

Let common difference be $$d.$$

First term \begin{align}a = 2 \end{align}

\begin{align} \text { Third term } &=a+2d \\ &=2+2d \end{align}

\begin{align} \text { Third term } &=26(\text { Given }) \\ 2+2d&=26 \\ 2d&=26-2 \\ 2d&=24 \\ d&=12 \end{align}

\begin{align} \text { Second term } &=a+d \\ &=2+12 \\ &=14 \end{align}

The missing term in the box is $$14.$$

ii) What is Known?

The AP with missing terms.

What is Unknown?

First and the third term.

Steps:

\begin{align}{a_n} = a + (n - 1)d\\{a_2} = 13,\,\,\,{a_4} = 3\end{align}

Second term $$= 13$$

\begin{align}a + (2 - 1)d &= 13\\a + d &= 13 \qquad\ldots (1)\end{align}

Fourth term $$= 3$$

$$\begin{array}{l} a + (4 - 1)d = 3\\ a + 3d = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots (2) \end{array}$$

Solving $$(1)$$ and $$(2)$$ for $$a$$ and $$d$$

Equation $$(2)$$ – Equation $$(1)$$ gives

\begin{align}{3d-d}&=3-13 \\ {2d}&=-10 \\ {d}&=-5\end{align}

Putting $$d$$ in equation $$(1)$$

\begin{align}{a + (-5)}&=13 \\ {a-5}&=13 \\ {a} &= 18\end{align}

\begin{align} \text { Third term } &=18-10 \\ &=8 \end{align}

Hence the missing terms in the boxes are $$18$$ and $$8.$$

iii) What is Known?

The AP with missing terms.

What is Unknown?

Missing terms in the boxes.

Steps:

First term $$a = 5$$

Fourth term \begin{align}{a_4} = \frac{{19}}{2}\end{align}

\begin{align} \text { Fourth term }{a_4}&= a +(4-1) d \\ &=5+3 d \\ 5+3 d &=\frac{19}{2} \\ 6 d &=9 \\ d &=\frac{3}{2} \end{align}

\begin{align}{\text { Second term }{a_2}}&=a+d \\&=5+\frac{3}{2} \\ &=\frac{13}{2}\end{align}

\begin{align}{\text{Third term }{a_3}}&= a + 2d\\ &= 5 + 2 \times \frac{3}{2}\\ &= 5 + 3\\ &= 8\end{align}

The missing terms in the boxes are \begin{align}\frac{{13}}{2}\end{align} and $$8.$$

iv) What is Known?

The AP with missing terms.

What is Unknown?

Missing terms in the boxes.

Steps:

First term \begin{align} a = -4 \end{align}

\begin{align} {\text { Sixth term } a_{6}}&=6 \\ a+5d &=6 \\-4+5d&=6 \\ 5d&=10 \\d&=2 \end{align}

\begin{align} {\text{Second term}}\!=\!a + d \!=\!\!-\! 4 + 2 =\,{\rm{ - 2}}\\{\text{Third term = a + 2d = }}{\rm{ - 4 + 4 = 0}}\\{\text{Fourth term = a + 3d = }}{\rm{ - 4 + 6 = 2}}\\{\text{Fifth term = a + 4d = }}\,\,{\rm{ - 4 + 8 = 4}} \end{align}

Hence the missing terms are \begin{align}-2,0,2,4. \end{align}

v) What is Known?

The AP with missing terms.

What is Unknown?

Missing terms in the boxes.

Steps:

Let the first term be \begin{align}= a. \end{align}

Common difference \begin{align}= d \end{align}

Second term \begin{align}= 38 \end{align} (Given)

\begin{align} a+d = 38 \quad…(1) \end{align}

Sixth term \begin{align} = -22 \end{align} (Given)

Sixth term \begin{align} = a + 5d = -22\quad \dots(2) \end{align}

Solving $$(1)$$ and $$(2)$$for $$a$$ and $$d$$

\begin{align}{4d}&= - 60 \\ {d}&=-15\end{align}

Put the $$d$$ value in …$$(1)$$

\begin{align}{a-15}&=38 \\ {a}&=53\end{align}

\begin{align} \text { Third term } &=a+2d \\ &=53-30=23 \end{align}

\begin{align} \text { Fourth term } &=a+3 d \\ &=53-45=8 \end{align}

\begin{align} \text { Fifth term } &=a+4d \\ &=53-60=-7 \end{align}

Hence the missing terms are \begin{align} 53, 23, 8, -7. \end{align}

## Chapter 5 Ex.5.2 Question 4

Which term of the

\begin{align} \text{AP: 3, 8, 13, 18……. is 78.}\end{align}

### Solution

What is Known?

The AP with missing terms

What is Unknown?

Which term will be $$78.$$

Reasoning:

\begin{align}a_{n}=a+(n-1) d \end{align}

Steps:

The AP is \begin{align} 3, 8, 13, 18. \end{align}

First term \begin{align} a &= 3 \end{align}

Second term \begin{align} a +d &= 8 \end{align}

Common difference \begin{align} d &= 8-3 = 5 \end{align}

\begin{align} a_{n}&= a+(n-1) d=78 \\ 3+(n-1) 5&=78 \\5(n-1)&=78-3 \\ n-1 &=15\\n&=16 \end{align}

$$78$$ is the $$16^\rm{th}$$ term of AP.

## Chapter 5 Ex.5.2 Question 5

Find the number of terms in each of the following APs:

i) \begin{align} 7,13,19,……,205\end{align}

ii) \begin{align}18,15 \frac{1}{2}, 13, \ldots . .,-47\end{align}

### Solution

What is Known?

The AP

What is Unknown?

No. of terms of the AP.

Reasoning:

\begin{align}{a_n} = a + \left( {n - 1} \right)d\end{align}

i) \begin{align} 7,13,19,……,205\end{align}

Steps:

\begin{align}{a_n} = a + \left( {n - 1} \right)d\end{align}

Where \begin{align}{a_n}\end{align} is the $$n^{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

\begin{align}{a}&=7 \\ {d}&=13-7=6 \\ {a_{n}}&=205 \\ {a_{n}} &=a+(n-1) d\\ {a+(n-1) d}&=205 \\ {7+(n-1) 6}&=205 \\ {(n-1) 6}&=205-7 \\ {(n-1) 6}&=198 \\ {n-1}&=33 \\ {n}&=33 + 1 \\ {n}&=34 \end{align}

Number of terms in the given AP is $$34.$$

ii) \begin{align}18,15 \frac{1}{2}, 13, \ldots . .,-47\end{align}

Steps:

\begin{align}d &= 15\frac{1}{2} - 18\\ &= \frac{{31}}{2} - 18\\&= {-\frac{{5}}{2}} \\{a_n} &= a + (n - 1)d\\a + (n - 1)d &= - 47\\18 + (n - 1)({-\frac{{5}}{2}}) &= - 47\\n - 1 &= 26\\n &= 26 + 1\\ &= 27 \end{align}

The number of terms in the given AP is $$27.$$

## Chapter 5 Ex.5.2 Question 6

Check Whether $$-150$$ is a term of the AP $$\,11,\,8,\,5,\,2\,\dots$$

### Solution

What is Known:?

The AP

What is Unknown?

Whether $$-150$$ is a term of AP.

Reasoning:

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

Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

$$11,\,8,\,5,\,2$$ are in AP

First term $$a = 11$$

Common difference $$d = 8--11 = - 3$$

\begin{align}{a_n} &= a + (n - 1)d\\a + (n - 1)d &= - 150\\11 + (n - 1)( - 3) &= - 150\\n - 1 &= \frac{{161}}{3}\\n &= \frac{{164}}{3}\end{align}

$$n = \frac{{164}}{3}$$ which is a fraction. Given number is not a term of AP $$11,\, 8,\, 5,\, 2\,\dots$$$$n$$ should be positive integer and not a fraction.

## Chapter 5 Ex.5.2 Question 7

Find the $$31^\rm{st}$$ term of an AP whose $$11^\rm{th}$$ term is $$38$$ and the $$16^\rm{th}$$ term is $$73.$$

### Solution

What is Known?

$$11^\rm{th}$$ and $$16^\rm{th}$$ term of AP.

What is Unknown?

$$31^\rm{st}$$ term of AP.

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

$\begin{array}{l}{a_n} = a + (n - 1)d\\{a_{11}} = 38\\a + (11 - 1)d = 38\\a + 10d = 38 \qquad \dots\left( 1 \right)\\{a_{16}} = 73\\a + 15d = 73\,\qquad \dots(2)\end{array}$

By solving the two equations (1) & (2) for $$a,d$$

\begin{align}5d &= 35\\d &= 7\end{align}

Putting $$d$$ in the (1) equation

\begin{align}a &= 38 - 70\\ &= - 32\end{align}

$$31^\rm{st}$$ terms is,

\begin{align}{a_{31}} &= a + (31 - 1)d\\ &= - 32 + 30 \times 7\\ &= - 32 + 210\\& = 178\end{align}

The $$31^\rm{st}$$ term of AP is $$178.$$

## Chapter 5 Ex.5.2 Question 8

An AP consists of $$50$$ terms of which $$3^\rm{rd}$$ term is $$12$$ and the last term is $$106.$$ Find the $$29^\rm{th}$$ term.

### Solution

What is Known:?

Number of terms in the AP.  $$3^\rm{rd}$$ term is $$12$$ and last term is $$106$$

What is Unknown?

$$29^\rm{th}$$ term.

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

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

Third term of AP is $$a + 2d$$

$a + 2d = 12 \qquad \dots(1)$

Last term $$= 106$$

$$50\rm{th}$$ term $$=106$$

\begin{align}a + \left( {50 - 1} \right)d &= 106\\a + 49d &= 106 \qquad \dots\left( 2 \right)\end{align}

Solving (1) & (2) for the values of $$a$$ and $$d$$.

\begin{align}47d &= 94\\d &= 2\end{align}

Putting $$d = 2$$ in equation (1)

\begin{align}a + 2 \times 2 &= 12\\a + 4 &= 12\\a &= 12 - 4\\a &= 8\end{align}

$$29^\rm{th}$$ term of AP is

\begin{align}{a_{29}} &= a + (29 - 1)d\\{a_{29}} &= 8 + \left( {28} \right)2\\{a_{29}} &= 8 + 56\\{a_{29}}& = 64\end{align}

$$29^\rm{th}$$ term of AP is $$64.$$

## Chapter 5 Ex.5.2 Question 9

If the $$3^\rm{rd}$$ and the $$9^\rm{th}$$ terms of an AP are $$4$$ and $$-8,$$ respectively, which term of this AP is zero.

### Solution

What is Known:?

$$3^\rm{rd}$$ and $$9^\rm{th}$$ term of AP,

What is Unknown?

Which term of AP is zero.

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Third term of the AP $$= 4$$

$a + 2d = 4 \qquad \dots\left( 1 \right)$

$$9^\rm{th}$$ term of AP $$= -8$$

$a + 8d = - 8 \qquad \ldots .\left( 2 \right)$

Solving (1) and (2) for $$a$$ and $$d$$

\begin{align}& \frac{\begin{align}& a\text{ }+\text{ }2d\text{ }=4 \\ & a\text{ }+\text{ }8d\text{ }=-8 \\ \end{align}}{-6d\text{ }=\text{ }12} \\ &\qquad \; d\text{ } =-2 \\\end{align}

Putting $$d = - 2$$ in equation (1)

\begin{align}a - 4 = 4\\a = 8\end{align}

\begin{align}a + (n - 1)d &= 0\\8 + (n - 1)( - 2)& = 0\\n - 1 &= 4\\n &= 5\end{align}

$$5^\rm{th}$$ term will be $$0.$$

## Chapter 5 Ex.5.2 Question 10

The $$17^\rm{th}$$ term of an AP exceeds its $$10^\rm{th}$$ term by $$7.$$ Find the common difference.

### Solution

What is Known:?

The difference between the $$17^\rm{th}$$ and $$10^\rm{th}$$ term.

What is Unknown?

Common difference $$d$$

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

\begin{align}{a_{17}} &= a + (17 - 1)d\\{a_{17}} &= a + 16d\\\\{a_{10}} &= a + (10 - 1)d\\{a_{10}} &= a + 9d\\\\{a_{17}} - {a_{10}} &= 7\\16d - 9d& = 7\\d &= 1\end{align}

The common difference is $$1.$$

## Chapter 5 Ex.5.2 Question 11

Which term of the AP $$3,\,15,\,27,\,39\dots$$ will be $$132$$ more than its $$54^\rm{th}$$ term?

### Solution

What is Known:?

The AP.

What is Unknown?

Which term will be $$132$$ more than $$54^\rm{th}$$ term.

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Given AP is $$3,\,15,\, 27,\, 39.$$

First term $$a = 3$$

Second term $$a + d = 15$$

$d = 15 - 3 = 12$

$$54^\rm{th}$$ term of the AP is

\begin{align}{a_{54}} &= a + (54 - 1)d\\ &= 3 + 53 \times 12\\ &= 3 + 636\\ &= 639\end{align}

Let $$n^\rm{th}$$ term of AP be $$132$$ more than $$54^\rm{th}$$ term (Given)

We get , $$132\,+\,639=771$$

${a_n} = 771$

\begin{align}{a_n} &= a + \left( {n - 1} \right)d\\771 &= 3 + \left( {n - 1} \right)12\\768 &= \left( {n - 1} \right)12\\\left( {n - 1} \right)& = 64\\n &= 65\end{align}

Therefore, $$65^\rm{th}$$ term will be $$132$$ more than $$54^\rm{th}$$ term.

Alternatively,

Let $$n^\rm{th}$$ term be $$132$$ more than $$54^\rm{th}$$ term.

\begin{align}n &= 54 + \frac{{132}}{{12}}\\& = 54 + 11 = {65^\rm{th}}\,\,{\rm{Term}}\end{align}

$$65^\rm{th}$$ term will be $$132$$ more than $$54^\rm{th}$$ term

## Chapter 5 Ex.5.2 Question 12

Two APs have the same common difference. The difference between their $$100^\rm{th}$$ term is $$100,$$ what is the difference between their $$1000^\rm{th}$$ terms?

### Solution

What is Known:?

Two APs with the same common difference and difference between their $$100^\rm{th}$$ term.

What is Unknown?

Difference between their $$1000^\rm{th}$$ term

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Let the first term of these A.P.s be $${a_1}$$ and $${b_1}$$ respectively and the Common difference of these A.P’s be $$d$$

For first A.P.,

\begin{align}{a_{100}} &= {a_1} + (100 - 1)d\\ &= {a_1} + 99{\rm{d}}\\{a_{1000}} &= {a_1} + (1000 - 1)d\\{a_{1000}} &= {a_1} + 999d\end{align}

For second A.P.,

\begin{align}{b_{100}} &= {b_1} + (100 - 1)d\\& = {b_1} + 99d\\{b_{1000}}& = {b_1} + (1000 - 1)d\\ &= {{\rm{b}}_1} + 999d\end{align}

Given that, difference between

$$100^\rm{th}$$ term of these A.P.s $$= 100$$

Thus, we have

\begin{align}&\left( {{a_1} + 99d} \right) - \left( {{b_1} + 99d} \right)= 100\\&{a_1} - {b_1}= 100\quad \dots {\rm{Equation}}\left( 1 \right)\end{align}

Difference between $$1000^\rm{th}$$ terms of these A.P.s

\begin{align}&\left( {{a_1} + 999d} \right)\! -\! \left( {{b_1}\! +\! 999d} \right) \\&=\! {a_1}\! - \!{b_1}\quad\dots{\rm{Equation}}\left( 2 \right)\end{align}

From equation (1) & Equation (2),

This difference, $${a_1} - {b_1} = 100$$

Hence, the difference between $$1000^\rm{th}$$ terms of these A.P. will be $$100.$$

## Chapter 5 Ex.5.2 Question 13

How many three-digit numbers are divisible by $$7$$?

### Solution

What is Known:?

Three-digit numbers

What is Unknown?

Number of all three-digit numbers which are divisible by $$7$$

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n \rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

First three-digit number that is divisible by $$7 = 105$$

Next number $$= 105 + 7 = 112$$

Therefore, the series becomes $$105,\, 112, \,119,\,\dots$$

All are three-digit numbers which are divisible by $$7$$ and thus, all these are terms of an A.P. having first term as $$105$$ and common difference as $$7.$$

When we divide $$999$$ by $$7,$$ the remainder will be $$5.$$

Clearly, $$999 − 5 = 994$$ is the maximum possible three-digit number that is divisible by $$7.$$

Hence the final series is as follows:

$105, \,112,\, 119,\, \dots, 994$

Let $$994$$ be the $$n\rm{th}$$ term of this A.P.

\begin{align}a &= 105\\d &= 7\\an &= 994\\n &= ?\end{align}

We know that the nth term of an A.P. Series,

\begin{align}{a_n} &= a + (n - 1)d\\994 &= 105 + (n - 1)7\\ 889 &= (n - 1)7\\n - 1 &= \frac{{889}}{7}\\n - 1& = 127\\{n}& = 127 + 1\\n &= 128\end{align}

Therefore, $$128$$ three-digit numbers are divisible by $$7.$$

## Chapter 5 Ex.5.2 Question 14

How many multiples of $$4$$ lie between $$10$$ and $$250$$?

### Solution

What is Known?

Numbers between $$10$$ and $$250$$

What is Unknown?

Multiples of $$4$$ between $$10$$ and $$250.$$

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

By Observation, First multiple of $$4$$ that is greater than $$10$$ is $$12.$$

Next will be $$16.$$

Therefore, the series will be as follows: $$12,\, 16,\, 20,\, 24,\, \dots$$

All these are divisible by $$4$$ and thus, all these are terms of an A.P. with first term as $$12$$ and common difference as $$4.$$

When we divide $$250$$ by $$4,$$ the remainder will be $$2.$$ Therefore, $$250 – 2 = 248$$ is divisible by $$4$$ which is the largest multiple of $$4$$ within $$250.$$

Hence the final series is as follows:

$12,\, 16,\, 20,\, 24,\, \dots, 248$

Let $$248$$ be the $$n\rm{th}$$ term of this A.P.

We know that the $$n\rm{th}$$ term of an A.P. Series,

\begin{align}a &= 12\\d &= 4\\{a_n} &= 248\\{a_n} &= a + (n - 1)d\\248 &= 12 + (n - 1)4\\\frac{{236}}{4} &= n - 1\\n& = 60\end{align}

Therefore, there are $$60$$ multiples of $$4$$ between $$10$$ and $$250.$$

## Chapter 5 Ex.5.2 Question 15

For what value of n, are the nth terms of two APs $$63,\, 65,\, 67,\,$$and $$3,\, 10,\, 17,\,\dots$$ equal?

### Solution

What is Known:?

Two different APs

What is Unknown?

Value of $$n$$ so that two APs have equal $$n^\rm{th}$$ term

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n^\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

If $$n\rm{th}$$ terms of the two APs $$63, \,65,\, 67,\dots$$ and $$3,\, 10,\, 17,\, \dots$$ are equal.

Then,

\begin{align} 63 + \left( {n-1} \right)2 =& 3 + \left( {n-1} \right)7 \\& \, \dots \text{Equation (1)}\end{align}

[Since In $$1\rm{st}$$ AP, $$a = 63$$, $$d = 65--3 = 2$$ and in $$2\rm{nd}$$ AP , $$a = 3$$,$$d = 10--3 = 7$$]

By Simplifying Equation (1)

\begin{align}7\left( {n-1} \right)-2\left( {n-1} \right)&= 63-3\\7n - 7 - 2n + 2 &= 60\\5n - 5 &= 60\\n &= \frac{{65}}{5}\\n &= 13\end{align}

The $$13^\rm{th}$$ terms of the two given APs are equal.

## Chapter 5 Ex.5.2 Question 16

Determine the A.P. whose third term is $$16$$ and the $$7^\rm{th}$$ term exceeds the $$5^\rm{th}$$ term by $$12.$$

### Solution

What is Known:?

Third term and relation between $$5^\rm{th}$$ term and $$7^\rm{th}$$ term.

What is Unknown?

The AP.

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Let $$a$$ be the first term and $$d$$ the common difference.

Hence from given, $${a_3} = 16$$ and $${a_7} - {a_5} = 12$$

\begin{align} a+\left( 3-1 \right)d&=16 \\ a+2d&=16 \quad \dots\left( 1 \right) \\\end{align}

Using $${a_7} - {a_5} = 12$$

\begin{align}\left[ {a + \left( {7 - 1} \right)d} \right] - \left[ {a + \left( {5 - 1} \right)d} \right] &= 12\\ [a + 6d] - [a + 4d] &= 12\\2d &= 12\\d &= 6\end{align}

By Substituting this in Equation (1), we obtain

\begin{align}a + 2 \times 6 &= 16\\a + 12 &= 16\\a &= 4\end{align}

Therefore, A.P. will be $$4+6,\,\,4+2\times 6,\,\,4+3\times 6\,\dots$$

Hence the series will be $$4, \,10,\, 16,\, 22,\,\dots.$$

## Chapter 5 Ex.5.2 Question 17

Find the $$20^\rm{th}$$ term from the last term of the A.P. $$3, \,8,\, 13,\, \dots 253$$

### Solution

What is Known:?

The AP

What is Unknown?

$$20^\rm{th}$$ term from the last term of AP.

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Given A.P. is $$3,\, 8,\, 13,\, \dots,253$$

From Given,

As the $$20^\rm{th}$$ term is considered from last $$a = 253$$

Common difference, $$d = 3 - 8 = - 5$$ (Considered in reverse order)

We know that the $$n\rm{th}$$ term of an A.P. Series,

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

Hence $$20^\rm{th}$$ Term, $${a_{20}} = a + (20 - 1)d$$

\begin{align}{a_{20}} &= 253 + (20 - 1)( - 5)\\& = 253 - 19 \times 5\\ &= 253 - 95\\& = 158\end{align}

Therefore, $$20^\rm{th}$$ term from the last term is $$158.$$

## Chapter 5 Ex.5.2 Question 18

The sum of $$4^\rm{th}$$ and $$8^\rm{th}$$ terms of an A.P. is $$24$$ and the sum of the $$6^\rm{th}$$ and $$10^\rm{th}$$ terms is $$44.$$ Find the first three terms of the A.P.

### Solution

What is Known:?

Sum of $$4^\rm{th}$$ and $$8^\rm{th}$$ terms is $$24$$ and sum of $$6^\rm{th}$$ and $$10^\rm{th}$$ terms is $$44.$$

What is Unknown?

First three terms of the AP.

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Let $$a$$ be the first term and $$d$$ the common difference.

Given,

\begin{align}{a_4} + {\rm{ }}{a_8}& = {\rm{ }}24\\\left( {a + 3d} \right) + \left( {a + 7d} \right) &= 24\\ \Rightarrow 2a + 10d &= 24\\ \Rightarrow a + 5d{\rm{ }} &= 12 \dots{\left( 1 \right)}\end{align}

Also,

\begin{align}{{a}_{6}}+{{a}_{10}}&=44 \\\left( a+5d \right)+\left( a+9d \right)&=44 \\\Rightarrow 2a+14d&=44 \\\Rightarrow a+7d&=22 \dots\left( 2 \right) \\\end{align}

On subtracting Equation (1) from (2), we obtain

\begin{align}(a + 7d)-\left( {a + 5d} \right) &= 22 - 12\\a + 7d - a - 5d &= 10\\2d &= 10\\d &= 5\end{align}

By Substituting the value of $$d = 5$$ in Equation (1), we obtain

\begin{align}a + 5d &= 12\\a + 5 \times 5 &= 12\\a + 25 &= 12\\a &= - 13\end{align}

The first three terms are $$a$$, $$\left( {a + d} \right)$$ and $$\left( {a + 2d} \right)$$

Substituting the values of $$a$$ and $$d$$, we get $$–13, (–13+5)$$ and $$(–13+2×5)$$

i.e., $$–13, –8$$ and $$–3$$

Therefore, the first three terms of this A.P. are $$−13, −8,$$ and $$−3.$$

## Chapter 5 Ex.5.2 Question 19

Subba Rao started work in $$1995$$ at an annual salary of $$\rm{Rs}\, 5000$$ and received an increment of $$\rm{Rs}\,200$$ each year. In which year did his income reach $$\rm{Rs}\,7000$$?

### Solution

What is Known:?

Annual salary of $$\rm{Rs}\,5000$$ and increment of $$\rm{Rs}\,200$$ each year.

What is Unknown?

The year in which income reached $$\rm{Rs}\,7000$$

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$nth$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

From the Given Data, incomes received by Subba Rao in the years $$1995,\,1996,\,1997\,\dots$$ are $$5000,\,5200, \,5400,\,\dots 7,000$$

From Observation,

\begin{align}a&= 5000\\d &= 200\end{align}

Let after $$n^\rm{th}$$ year, his salary be $$\rm{Rs}\, 7000.$$

Hence $${a_n} = 7000$$

We know that the $$n^\rm{th}$$ term of an A.P. Series,

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

By Substituting above values,

\begin{align}7000 &= 5000 + (n - 1)200\\200(n - 1) &= 7000 - 5000\\n - 1 &= \frac{{2000}}{{200}}\\ n &= 10 + 1\\n &= 11\end{align}

Therefore, in $$11^\rm{th}$$ year, his income reached $$\rm{Rs}\,7000.$$ Which means after $$10$$ years of $$1995$$ i.e. $$1995 + 10 \Rightarrow 2005$$

In $$2005$$ his income reached $$\rm{Rs}\,7000$$.

## Chapter 5 Ex.5.2 Question 20

Ramkali saved $$\rm{Rs}\, 5$$ in the first week of a year and then increased her weekly saving by $$\rm{Rs}\, 1.75.$$ If in the $$n^\rm{th}$$ week, her weekly savings become $$\rm{Rs}\, 20.75,$$ find $$n.$$

### Solution

What is Known:?

Savings in first week $$\rm{Rs}\, 5$$ and increment of $$\rm{Rs}\, 1.75$$ weekly in savings.

What is Unknown?

Week in which her savings become $$\rm{Rs}\, 20.75$$

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n^\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

From the given data, Ramkali’s savings in the consecutive weeks are

$$\rm{Rs}\, 5, \,\rm{Rs}\, (5+1.75),\, \rm{Rs}\, (5+2×1.75), \\\rm{Rs}\, (5+3×1.75)\, \dots$$ and so on

Hence in $$n^\rm{th}$$ weeks savings, $$\rm{Rs}\, [5+(n−1)×1.75] = \rm{Rs}\, 20.75$$

Now from the above we know that

\begin{align} a&=5 \\ d&=1.75 \\{{a}_{n}}&=20.75 \\n&=?\end{align}

We know that the $$n^\rm{th}$$ term of an A.P. Series,

\begin{align}{a_n} &= a + (n - 1)d\\20.75& = 5 + (n - 1)1.75\\15.75 &= (n - 1)1.75\n - 1)& = \frac{{15.75}}{{1.75}}\\n - 1 &= \frac{{1575}}{{175}}\\n - 1 &= 9\\n &= 10\end{align} The answer is \(n = 10

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