# NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.2

Arithmetic Progressions

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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video Solution**

**What is Known:?**

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

**What is Unknown?**

Common difference \(d\)

**Reasoning: **

**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**

**Video Solution**

**What is Known:?**

The AP**. **

**What is Unknown?**

Which term will be \(132\) more than \(54^\rm{th}\)^{ }term**. **

**Reasoning: **

**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**

**Video 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: **

**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**

**Video 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**

**Video Solution**

**What is Known?**

Numbers between \(10\) and \(250\)

**What is Unknown?**

Multiples of \(4\) between \(10\) and \(250.\)

**Reasoning: **

**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**

**Video 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**

**Video Solution**

**What is Known:?**

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

**What is Unknown?**

The AP.

**Reasoning: **

**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**

**Video Solution**

**What is Known:?**

The AP

**What is Unknown?**

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

**Reasoning: **

**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**

**Video 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: **

**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**

**Video 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**

**Video 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 \(n = 10\)