NCERT Exemplar Class 10 Maths Solutions Chapter 5 Arithmetic Progressions

# NCERT Exemplar Class 10 Maths Solutions Chapter 5 Arithmetic Progressions

Edited By Ravindra Pindel | Updated on Sep 05, 2022 03:26 PM IST | #CBSE Class 10th

NCERT exemplar Class 10 Maths solutions chapter 5 provides the understanding for arithmetic progressions. Arithmetic Progressions is a very useful concept and can be used as a tool to generate useful value from a given set of data. The NCERT exemplar Class 10 Maths chapter 5 solutions are prepared by our highly experienced mathematics department and provide all the necessary in-depth concepts required to study NCERT Solutions for Class 10 Maths. These NCERT exemplar Class 10 Maths chapter 5 solutions take into consideration all the basic concepts of arithmetic progression so that students can build a strong foundation. The CBSE Syllabus for Class 10 is incorporated in these NCERT exemplar Class 10 Maths solutions chapter 5.

## Question:1

In an AP, if d = –4, n = 7, an=4, then a is

(A) 6

(B) 7

(C) 20

(D) 28

Given
$\\a\textsubscript{n} = 4, d = - 4, n = 7 \\Use \ a\textsubscript{n} = a + (n - 1).d\\ a + (7 - 1). (- 4) = 4\\ a +(6).(-4) = 4\\ a - 24 = 4\\ a = 4 + 24\\ a = 28\\$
Hence option D is correct

## Question:1

In an AP, if a = 3.5, d = 0, n = 101, then an will be
(A) 0
(B) 3.5
(C) 103.5
(D) 104.5

We know that
$\\ a\textsubscript{n} = a + (n -1).d\\ Given, a = 3.5, d = 0, n = 101\\ a\textsubscript{n} = a + (n - 1) d\\ a\textsubscript{n} = 3.5 + (101 - 1).0\\ a\textsubscript{n} = 3.5 + 0\\ a\textsubscript{n} = 3.5\\$
Hence option B is correct.

Question:3

The list of numbers – 10, – 6, – 2, 2... is
(A) an AP with d = – 16
(B) an AP with d = 4
(C) an AP with d = – 4
(D) not an AP

$\\a\textsubscript{1}= - 10, a\textsubscript{2} = - 6, a\textsubscript{3} = -2, a\textsubscript{4} = 2\\ a\textsubscript{2} - a\textsubscript{1 }= - 6 - (-10)\\ \ \ \ \ \ \ \ \ \ \ \ = - 6 + 10 \\ = 4\\ a\textsubscript{3} - a\textsubscript{2 }= -2 - (-6)\\ \ \ \ \ \ \ \ \ \ \ = - 2 + 6 \\ \ \ \ \ \ \ \ \ \ \ = 4\\ a\textsubscript{4} - a\textsubscript{3 }= 2 - (-2)\\ \ \ \ \ \ \ \ \ \ \ = 2 + 2\\ \ \ \ \ \ \ \ \ \ \ = 4\\$
Hence the given sequence is an AP with d = 4
ANS - (B)

Question:4

Here AP is
$-5, -\frac{5}{2},0,\frac{5}{2}, \ldots \ldots ..\\$
Here
\\a\textsubscript{1 }= - 5, a\textsubscript{2}= -\frac{5}{2} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Common difference) d = a\textsubscript{2} - a\textsubscript{1\ \ \ }\\ = -\frac{5}{2}-(-5) \\ = -\frac{5}{2}+5 \\ \begin{aligned} & =\frac{-5+10}{2} \\ & =\frac{5}{2} \\ \end{aligned}
$\\ a \textsubscript{n} = a\textsubscript{1} + (n - 1).d\\ Put\ n = 11\\ a\textsubscript{11} = a\textsubscript{1} + (11 - 1).d \\ a\textsubscript{11} = - 5 + (10). (5/2)\\ \ \ \ \ = - 5 + 5.(5)\\ \ \ \ \ = - 5 + 25\\ \ \ \ = 20\\$
Hence option B is correct.

Question:5

The first four terms of an AP, whose first term is –2 and the common difference is –2, are
(A) – 2, 0, 2, 4
(B) – 2, 4, – 8, 16
(C) – 2, – 4, – 6, – 8
(D) – 2, – 4, – 8, –16

Given,
$\\a = - 2, d = - 2\\ a + d = - 2 + (-2)\\ = - 4\\ a + 2d = - 2 + 2(-2)\\ =-2 - 4 \\ \ \ = - 6\\ a + 3d = - 2 + 3(-2)\\ =-2 - 6 \\ = - 8\\$
Hence the given AP is -2, -4, -6, - 8
Hence option (C) is correct.

Question:6

The 21st term of the AP whose first two terms are –3 and 4 is
(A) 17
(B) 137
(C) 143
(D) –143

Given
$\\a\textsubscript{1} = - 3\\ a\textsubscript{2} = 4\\ (Common difference) d = a\textsubscript{2} - a\textsubscript{1}\\ \ = 4 - (-3)\\ \ = 4 + 3 \\ \ = 7 \\ We know that\\ \ a\textsubscript{n}\ =\ a\ +\ (n\ -\ 1)d \ \ put n = 21\\ a\textsubscript{21} = a + (21 - 1).d\\ \ \ \ \ = - 3 + (20).(7)\\ \ \ \ \ = - 3 + 140 \\ \ \ \ \ = 137\\$
Hence option B is correct.

Question:7

If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
(A) 30
(B) 33
(C) 37
(D) 38

We know that
$\\a\textsubscript{n }= \textsubscript{ }a + ( n - 1 ).d \ldots ..(1)\\ Given a\textsubscript{2} = 13\\ a\textsubscript{5} = 25\\ a + d = 13 \ldots \ (2)\ \ \ \ (Put n = 2 in eq. (1) )\\ a + 4d = 25 \ldots \ (3)\ \ \ \ \ (Put n = 5 in eq. (1) )\\ \ Subtract eq. (1) and (2)\\ - 3d = - 12\\ \ \ \ \ d = \frac{12}{3} \\ \ \ \ d = 4\\ Put d= 4 in equation (2)\\ a + 4 = 13\\ a = 13 - 4\\ a = 9\\ a\textsubscript{7}\ =\ a\ +\ (7\ -\ 1).d\ \ \ \ (Put n = 7 in eq. (1) )\\ \ \ = 9 + (6). (4)\\ \ = 9 + 24\\ = 33\\$
Hence option B is correct.

Question:8

Which term of the AP: 21, 42, 63, 84... is 210?
(A) 9th
(B) 10th
(C) 11th
(D) 12th

Here,
$\\a\textsubscript{n} = 210\\ \text{ AP is 21, 42, 63, 84} \ldots \\ Here\ \ a\textsubscript{1}=\ 21\ \ \ \ \ a\textsubscript{2}{=} 42\\ d = a\textsubscript{2} - a\textsubscript{1 \ \ \ \ \ \ \ \ \ \ \ \ }(d=common difference)\\ d= 42 - 21 \ \ \ (a\textsubscript{1}= first term) \\ \ \ d = 21\\ \ a\textsubscript{n} = 210\\ a\textsubscript{1}+ (n - 1).d = 210\\ 21 + (n - 1).21 = 210\\ 21+21n - 21 = 210\\ \ 21n = 210\\$
$\\ =\frac{210}{21} \\ \\ \ \ \ \ \ n = 10\\ \text{Hence} 10\textsuperscript{th} \text{term of the AP is 210}\\$

Question:9

Given
$\\ d = 5 \\ a\textsubscript{18 }- a\textsubscript{13 }= ?\\ \text{We know that }a\textsubscript{n} = a + (n - 1).d\\ Put n = 18, a\textsubscript{18} = a + (18 - 1).d\\ Put n = 13, a\textsubscript{13} = a + (13 - 1).d\\ Now a\textsubscript{18 }- a\textsubscript{13 }= (a + (18 - 1).d) - (a + (13 - 1).d)\\ = a + 17d - a - 12d\\ = 17d - 12d\\ = 5d\\ =5 \times 5\\ = 25\\$
Option is (C) is correct.

Question:10

Given,
$\\a\textsubscript{18 }- a\textsubscript{14} = 32 \\ We know that a\textsubscript{n} = a + (n - 1).d\\ Put n = 18\\ a\textsubscript{18} = a + (18 - 1).d\\ a\textsubscript{18} = 9 + 17d \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ..(1)\\ n = 14\\ a\textsubscript{14} = a + (14 - 1).d\\ a\textsubscript{14} = 9 + 13d \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .(2)\\ subtract eq (2) from eq(1)\\ a\textsubscript{18}- a\textsubscript{14} = 32\\ (a + 17d) - (a + 13d) = 32\\ a + 17d - a - 13d = 32\\ 4d = 32\\ d = \frac{32}{4} \\ d=8 \\$
Hence common difference is 8.

Question:11

$\\ \text{Let} \ first \ AP \ is \ a\textsubscript{1}, a\textsubscript{2}, a\textsubscript{3} \ldots \ldots \ldots .(1)\\ \text{ Second AP is } b\textsubscript{1}, b\textsubscript{2}, b\textsubscript{3} \ldots \ldots \ldots \ldots ..(2)\\ \text{Given,} a\textsubscript{1} = -1, b\textsubscript{1} = - 8\\ \text{ Difference between their } 4\textsuperscript{th} \ terms\ is \\ a_4 - b\textsubscript{4} = (a\textsubscript{1} + 3d) - (b\textsubscript{1}+3d)\\ \text{Given that common difference of both A.P are equal} \\ a_4 - b\textsubscript{4} \\= -1 - (-8)\\ = -1 + 8\\ = 7\\$

Hence option (C) is correct

Question:12

$\\ \text{ Given }7 \times (a\textsubscript{7}) = 11 \times (a\textsubscript{11})\\ 7 \times (a + 6d) = 11 \times (a+10d) (Using a\textsubscript{n} = a + (n - 1) \times d )\\ 7a + 42d = 11a + 110d\\ 11a-7a+110d-42d=0\\ 4a + 68d = 0\\ \text{Dividing by 4 we get}\\ a + 17d = 0 \ldots (1)\\ a\textsubscript{18} = a + (18 - 1)d\\ = a + 17d\\ = 0 \\$

Hence option D is correct.

Question:13

The 4thterm from the end of the AP: –11, –8, –5, ..., 49 is
(A) 37
(B) 40
(C) 43
(D) 58

$\\ \text{Given AP is -11, -8, -5} \ldots \ldots 49\\ \text{ Here} , a\textsubscript{1} = -11 a\textsubscript{2 }=-8 a\textsubscript{n}=49\\ d = a\textsubscript{2} - a\textsubscript{1}\\ = -8- (-11) \\ = -8 +11\\ = 3\\ \text{ We know that }a\textsubscript{n} = a\textsubscript{1} + (n - 1).d\\ \text{Put 49 }= -11 + (n - 1)(3) \\ 49 = -11 +3n -3 \\ 49 = - 14+3n \\ 49+14 = 3n\\ 63 = 3n\\ n=\frac{63}{3} \\ n=21 \\ a\textsubscript{n} = a\textsubscript{1} + (n - 1).d \\ Put n= 21-3=18\\ a\textsubscript{18}= -11 + (18 - 1) \times 3 \\ a\textsubscript{18}= -11 + 17 \times 3 \\ a\textsubscript{18}= -11 + 51\\ a\textsubscript{18}= 40 \\ \text{40 is the } 4\textsuperscript{th } \text{last term from end}\\$
Hence option B is correct

Question:14

The famous mathematician associated with finding the sum of the first 100 natural numbers is
(A) Pythagoras
(B) Newton
(C) Gauss
(D) Euclid

[C]
Famous mathematician gauss find the sum of the first 100 natural numbers.
According to which when he added the first and the last number in the series, the second and the second last number in the series, and so on. He get a sum of 101 and there are 50 pairs exist so the sum of 100 natural numbers is 50×101=5050.

Question:15

$\\Given a=-5,d=2 here a= first term and d= common difference\\ S\textsubscript{6} = \frac{6}{2}(2a+(6-1)d) \\ = 3(2 \times (-5) + 5 \times 2)\\ = 3(-10 + 10)\\ = 3(0) = 0\\$
Hence option A is correct

Question:16

The sum of first 16 terms of the AP: 10, 6, 2... is
(A) –320
(B) 320
(C) –352
(D) –400

$\\Given:AP=10,6,2 \ldots here a\textsubscript{1}=firsttermandd=commondifference \\ a\textsubscript{1} = 10, a\textsubscript{2}=6\\ d = a\textsubscript{2} - a\textsubscript{1 }\\ d =6 - 10\\ d = - 4\\ {{S}_{16}}=\frac{16}{2}(2a+(16-1)d) \\ = 8(2 \times 10 + 15 \times (-4))\\ = 8(20 - 60)\\ = 8 \times (-40) = - 320\\$
Hence option A is the correct answer.

Question:17

$\\Given : a = 1, a\textsubscript{n} = 20, S\textsubscript{n} = 399\\ {{S}_{n}}=\frac{n}{2}(a+{{a}_{n}}) \\ 38 = n\\$
Hence option C is correct.

Question:18

The sum of first five multiples of 3 is
(A) 45
(B) 55
(C) 65
(D) 75

$\\Given:- Multiple of 3 is : 3, 6, 9, 12, \ldots \ldots .\\ a\textsubscript{1 }= 3, d = a\textsubscript{2} - a\textsubscript{1}\\ d = 6 - 3\\ d = 3\\ {{S}_{5}}=\frac{5}{2}[2{{a}_{1}}+(5-1)d] \\ =\frac{5}{2}(2\times 3+4\times 3) \\ =\frac{5}{2}(6+12) \\ =\frac{5}{2}\times (18) \\$
Hence option A is correct.

## Question:1

–1, –1, –1, –1, ...
ii)
0, 2, 0, 2, ...
iii)
1, 1, 2, 2, 3, 3,...
iv)
11, 22, 33,...
v)
$\frac{1}{2}, \frac{1}{3}, \frac{1}{4},......................$
vi)
$2, 2^2, 2^3, 2^4, ...$
vii)
$\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots$

i)
Solution.
$\\-1, -1, -1, -1 ... \text{here} \ a\textsubscript{1 }= -1, \ a\textsubscript{2 }= -1, \ a\textsubscript{3 }= -1, a\textsubscript{4}= -1\\ a\textsubscript{2} - a\textsubscript{1} = - 1 - (-1) = - 1 + 1 = 0\\ a\textsubscript{3} - a\textsubscript{2} = + 1(-1) = -1 + 1 = 0\\ a\textsubscript{4 }- a\textsubscript{3} = - 1 - (-1) = - 1 + 1 = 0\\$
Here the difference of the successive term is same hence it is an AP.
ii)
Solution.
$\\ \text{The given series is 0, 2, 0, 2 ... here} \ a\textsubscript{1 }= 0, \ a\textsubscript{2 }= 2, \ a\textsubscript{3 }= 0 , \ a\textsubscript{4}= 2 \\ a\textsubscript{2} - a\textsubscript{1} = 2 - 0 = 2\\ a\textsubscript{3} - a\textsubscript{2} = 0 - 2 = - 2\\ a\textsubscript{4} - a\textsubscript{3} = 2 - 0 = 2\\$
Here the difference successive term is not same. Hence it is not an AP.
iii)
Solution.
$\\ \text{The given series is }{1, 1, 2, 2, 3, 3,...} \ here \ a\textsubscript{1 }= 1, a\textsubscript{2 }= 1, a\textsubscript{3 }= 2, a\textsubscript{4}= 2\\ a\textsubscript{2} - a\textsubscript{1} = 1 - 1 = 0\\ a\textsubscript{3} - a\textsubscript{2} = 2 - 1 = - 1\\ a\textsubscript{4} - a\textsubscript{3} = 2 - 2 = 0\\$

Here the difference successive term is not the same. Hence it is not an AP.
iv)
Solution.
$\\ \text{The given series is } {11, 22, 33,...} \text{here } a\textsubscript{1 }= 11, a\textsubscript{2 }= 22, a\textsubscript{3 }= 33\\ a\textsubscript{2} - a\textsubscript{1} = 22 - 11 = 11\\ a\textsubscript{3} - a\textsubscript{2} = 33 - 22 = 11\\$

Here the difference of the successive term is same. Hence it is an AP.
v)
Solution.
\begin{aligned} &\text { Here the given series is } \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots \ldots \ldots \text { here } \mathrm{a}_{1}=\frac{1}{2} \mathrm{a}_{2}=\frac{1}{3} \mathrm{a}_{3}=\frac{1}{4}\\ &\mathrm{a}_{2}-\mathrm{a}_{1}=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=\frac{-1}{6}\\ &a_{3}-a_{2}=\frac{1}{4}-\frac{1}{3}=\frac{3-4}{12}=\frac{-1}{12} \end{aligned}
Here the difference between the successive terms is not the same. Hence it is not an AP.
vi)
Solution.
$\\ \text{Here the given series is }2, 2\textsuperscript{2}, 2\textsuperscript{3}, 2\textsuperscript{4} \ldots \\ \text{here }a\textsubscript{1 }= 2 a\textsubscript{2 }= 2\textsuperscript{2 },a\textsubscript{3 }= 2\textsuperscript{3 } a\textsubscript{4}= 2\textsuperscript{4}\\ a\textsubscript{2} - a\textsubscript{1} = 2\textsuperscript{2} - 2 = 4 - 2 = 2\\ a\textsubscript{3} - a\textsubscript{2} = 2\textsuperscript{3} - 2\textsuperscript{2} = 8 - 4 = 4\\ a\textsubscript{4} - a\textsubscript{3} = 2\textsuperscript{4} - 2\textsuperscript{3} = 16 - 8 = 8\\$

Here the difference of the successive terms is not the same. Hence it is not an AP.
vii)
Solution.

\begin{aligned} &\text { Here the given series is } \sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots . \text { here } \mathrm{a}_{1}=\sqrt{3} \mathrm{a}_{2}=\sqrt{12} \mathrm{a}_{3}=\sqrt{27}\\ &a_{4}=\sqrt{48}\\ &a_{2}-a_{1}=\sqrt{12}-\sqrt{3}=2 \sqrt{3}-\sqrt{3}=\sqrt{3}=\sqrt{3}(2-1)=\sqrt{3}\\ &a_{3}-a_{2}=\sqrt{27}-\sqrt{12}=3 \sqrt{3}-2 \sqrt{3}=\sqrt{3}\\ &a_{4}-a_{3}=\sqrt{48}-\sqrt{27}=4 \sqrt{3}-3 \sqrt{3}=\sqrt{3} \end{aligned}
Here the difference of the successive term is the same. Hence it is an AP.

Question:2

Justify whether it is true to say that $-1,-\frac{3}{2},-2, \frac{5}{2}, \ldots$ forms an A.P.as $a_2 - a_1 = a_3 - a_2.$

Solution.

$-1,-\frac{3}{2},-2, \frac{5}{2}, \ldots \ldots \\ \text{here} \ \ \mathrm{a}_{1}=-1 \ \ \mathrm{a}_{2}=-\frac{3}{2} \ \ \mathrm{a}_{3}=-2 \ \ \mathrm{a}_{4}=\frac{5}{2}$
$\\ a_{2}-a_{1}=-\frac{3}{2}-(-1)=\frac{-3}{2}+1=\frac{-3+2}{2}=\frac{-1}{2} \\\\ a_{3}-a_{2}=-2-\left(\frac{-3}{2}\right)=-2+\frac{3}{2}=\frac{-4+3}{2}=\frac{-1}{2}$
$\text{Here}$ $a_{2}-a_{1}=a_{3}-a_{2},$ $\text{Hence it is an A P }$

Question:3

For the AP: –3, –7, –11, ..., can we find directly $a_{30} - a_{20}$ without actually finding $a_{30} and a_{20}$?

Solution. Here the given series is
$-3,-7,-11 \ldots$
$a_{1}=-3, a_{2}=-7$
$\\ \mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1} \\ =-7-(3) \\ =-7+3 \\ =-4 \\ \mathrm{a}_{30}-\mathrm{a}_{20}=\mathrm{a}+(30-1) \mathrm{d}-(\mathrm{a}+(20-1) \mathrm{d}) \\ \quad=\mathrm{a}+29 \mathrm{~d}-\mathrm{a}-19 \mathrm{~d} \\ =10 \mathrm{~d} \\ =10 \times(-4) \\ =-40$
$\text{Yes we can find}$ $\mathrm{a}_{30}-\mathrm{a}_{20}$ $\text{without actually calculating}$ $\mathrm{a}_{30} \ and \ \mathrm{a}_{20} .$

Question:4

Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21stterms, which is the same as the difference between any two corresponding terms. Why?

Solution.
It is given that the first term of an AP is 2 and of other is 7.
$\\ \text{ Let}\\ a\textsubscript{1} = 2 b\textsubscript{1} = 7\\ \text{Common difference of both A.P is same} \\ A\textsubscript{n} \text{ is nth term for first A.P},b\textsubscript{n} \text{ is nth term of second A.P}\\ \text{As per the question}\\ a\textsubscript{10} - b\textsubscript{10} = a\textsubscript{1}+ 9d - b\textsubscript{1} - 9d\\ = a\textsubscript{1}- b\textsubscript{1} = 2 - 7 = - 5 \\ a\textsubscript{21 }- b\textsubscript{21} = a\textsubscript{1} + 20d - b\textsubscript{1} - 20d\\ = a\textsubscript{1} - b\textsubscript{1 }= - 5 \\ \text{Difference between first term = }a\textsubscript{1} - b\textsubscript{1 }= - 5\\$
it is because when we find the difference between 10th and 21st term it is equal to the a1-b1 which is the difference between first terms.

Question:5

Is 0 a term of the AP: 31, 28, 25, ...? Justify your answer.

Solution.
Here he given series is 31, 28, 25, …….
$\\a = 31\\ d = a\textsubscript{2}- a\textsubscript{1} = 28 - 31= - 3\\ If 0 is a term of this A.P then value of n must be a positive integer.\\ a\textsubscript{n} = a + (n - 1)d\\ 0 = a + (n - 1)d\\ 0 = 31 + (n - 1) (-3)\\ 0 = 31 - 3n + 3\\ 3n = 34\\ n=\frac{34}{3}=11.333. \\$
Which is not an integer.
Hence 0 is not a term of this AP.

Question:6

The taxi fare after each km, when the fare is Rs 15 for the first km and Rs. 8 for each additional km, does not form an AP as the total fare (in Rs) after each km is
15, 8, 8, 8, ...
Is the statement true? Give reasons.

Solution.
$\\ \text{ Given series is = 15, 8, 8, 8,} \ldots \ldots here a\textsubscript{1 }=15 a\textsubscript{2 }= 8, a\textsubscript{3 }= 8, a\textsubscript{4}= 8\\ a\textsubscript{2} - a\textsubscript{1} = 8 - 15 = - 7\\ a\textsubscript{3}- a\textsubscript{2} = 8 - 8 = 0\\ a\textsubscript{4}- a\textsubscript{3} = 8 - 8 = 0\\$
Here difference of the successive terms is not same hence it does not form an AP
Therefore given statement is true.

Question:7

i)In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.
The fee charged from a student every month by a school for the whole session, when the monthly fee is Rs 400.
ii)
In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.
The fee charged every month by a school from Classes I to XII, when the monthly fee for Class I is Rs 250, and it increases by Rs 50 for the next higher class.
iii)
In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.
The amount of money in the account of Varun at the end of every year when Rs 1000 is deposited at simple interest of 10% per annum.
iv)
In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.
The number of bacteria in a certain food item after each second, when they double in every second.

i)
Solution.
According to the question:
It is clear that the monthly fee charged from students is = 400 Rs.
Hence series is ; 400, 400, 400, 400 ……
the difference of the successive terms is same so It forms an AP with common difference d = 400 – 400 = 0
ii)
Solution.
According to the question:
Fees started Rs. 250 and increased by 50 Rs. from I to XII is
$\\\text{Here first term}( a\textsubscript{1})=250 common difference (d)=50 \\ a\textsubscript{1}=250\\ a\textsubscript{2}=250+50\\ =300\\ a\textsubscript{3}=300+50\\ =350\\ a\textsubscript{4}=350+50\\ =400\\ So the A.P. 250, 300, 350, 400 \ldots \\$
It forms an AP with common difference 50.
iii)
Solution. We know that: Simple interest = (P × R × T)/100
$\\ \text{ Here P = 1000}Rs \\ R=10 \% P.a\\ T=1 year \\ S.I= \frac{1000\times 10\times 1}{100} \\ = 100\\ \text{Here common difference (d)}=100 Rs and first term(a\textsubscript{1}) = 1000 Rs\\ a\textsubscript{1}=1000 Rs\\ a\textsubscript{2}=1000+100\\ =1100 Rs\\ a\textsubscript{3}=1000+200\\ =1200 Rs\\ a\textsubscript{4}=1000+300\\ =1300 Rs\\ \text{So the} A.P: 1000, 1100, 1200, 1300 \ldots \\$
This is definite form an AP with common difference of SI=100 Rs.
iv)
Solution.
Let the number of bacteria = x
According to the question:
They double in every second.
$\\ \therefore x, 2x, 2(2x), 2(2 \times 2x) \ldots \ldots \ldots \ldots \ldots \\ Here a\textsubscript{1 }= x , a\textsubscript{2 }=2x, a\textsubscript{3 }=4x, a\textsubscript{4}=8x\\ x, 2x, 4x, 8x \ldots .. \ldots ..\\ a\textsubscript{2} - a\textsubscript{1} = 2x - x = x\\ a\textsubscript{3}- a\textsubscript{2} = 4x - 2x = 2x\\$
Since the difference between the successive terms is not same therefore it does not form an AP.

Question:8

i)
Justify whether it is true to say that the following are the nth terms of an AP.
2n–3
ii)
Justify whether it is true to say that the following are the nth terms of an AP.
$3n^2+5$
iii)
Justify whether it is true to say that the following are the nth terms of an AP.
$1+n+n^2$

i)

Solution.

$\\\text{Here} a\textsubscript{n} = 2n - 3\\ Put n = 1, a\textsubscript{1} =2(1) - 3 = - 1\\ Putn=2, a\textsubscript{2} =2(2) - 3 = 4 - 3 = 1\\ Putn=3, a\textsubscript{3}=2(3) - 3 = 6 - 3 = 3\\ Putn=4, a\textsubscript{4} =2(4) - 3 = 8 - 3 = 5\\ Numbers are : -1, 1, 3, 5, \ldots ..\\ a\textsubscript{2}- a\textsubscript{1} = 1 - (-1) = 1 + 1 = 2\\ a\textsubscript{3}- a\textsubscript{2} = 3 - 1 = 2\\ a\textsubscript{4}- a\textsubscript{3} = 5 - 3 = 2\\$
Hence, 2n – 3 is the nth term of an AP because the common difference is the same.
ii)
Solution.
$\\\text{Here, }a\textsubscript{n} = 3n\textsuperscript{2} + 5\\ Put n = 1, a\textsubscript{1} =3(1)\textsuperscript{2} + 5 = 3 + 5 = 8\\ Put n =2, a\textsubscript{2} =3(2)\textsuperscript{2} + 5 = 12 + 5 = 17\\ Put n = 3, a\textsubscript{3} =3(3)\textsuperscript{2} + 5 = 27 + 5 = 32\\ Numbers are: 8, 17, 32 \ldots \\ a\textsubscript{2} - a\textsubscript{1} = 17 - 8 = 9\\ a\textsubscript{3}- a\textsubscript{2} = 32 - 17 = 15\\$
Here the common difference is not the same
Therefore series $3n^2+5$ is not the nth term of an AP.

iii)
Solution.
$\\\text{Here} a\textsubscript{n} = 1 + n + n\textsuperscript{2}\\ Put n = 1, a\textsubscript{1} = 1 + 1 + (1)\textsuperscript{2} = 3\\ Put n = 2, a\textsubscript{2} = 1 + 2 + (2)\textsuperscript{2} = 7\\ Put n = 3, a\textsubscript{3} = 1 + 3 + (3)\textsuperscript{2} = 13\\ The numbers are 3, 7, 13 \ldots \\ a\textsubscript{2} - a\textsubscript{1} = 7 - 3 = 4\\ a\textsubscript{3}- a\textsubscript{2} = 13 - 7 = 6\\$
Here the common difference is not the same
$\text{Hence}$ $1 + n + n^2$ $\text{is not the nth term of an AP.}$

## Question:1

Match the APs given in column A with suitable common differences given in column B.

 Column A Column B (A1) 2, – 2, – 6, – 10,... (B1) $\frac{2}{3}$ (A2) a = –18, n = 10, an = 0 (B2) –5 (A3) a = 0, $a_{10}$ = 6 (B3) 4 (A4) $a_2 = 13, a_4 =3$ (B4) –4 (B5) 2 (B6)$\frac{1}{2}$ (B7) 5

$\\(A\textsubscript{1}) \: \: \: \: \\2, -2, -6, -10, \ldots ..\\ a\textsubscript{2}= -2, a\textsubscript{1}= 2\\ d = a\textsubscript{2} - a\textsubscript{1}\\ -2 - 2 = - 4 \: \: \: \: (B\textsubscript{4}) - 4\\ (A\textsubscript{2}) \\a = - 18, n = 10, a\textsubscript{n} = 0\\ a\textsubscript{n} = 0\\ \because a + (n - 1) d = 0\\ -18 + (10 - 1)d = 0\\ 9d = 18\\ \vspace{\baselineskip} d = 2 \: \: \: \: (B\textsubscript{5}) 2\\ (A\textsubscript{3})\\ a = 0, a\textsubscript{10} = 6\\ a + 9d = 6\\ 0 + 9d = 6\\ d=\frac{6}{9}=\frac{2}{3} \: \: \: \: (B\textsubscript{1}) 2/3\\ (A\textsubscript{4})\\ a\textsubscript{2} = 13, a\textsubscript{4} = 3\\ \because a + (n - 1) d = 0\\ a+d=13 \\ a + 3d=3 \\ - \_-\_\_\_-\_\_\_\_\_\_\\ -2d = 10\\ d=\frac{10}{-2} \\ d = - 5 \: \: \: \: (B\textsubscript{2}) - 5\\$

Question:2

i)

Verify that each of the following is an AP, and then write its next three terms.
$0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots$
ii)
Verify that each of the following is an AP, and then write its next three terms.
$5, \frac{14}{3}, \frac{13}{3}, 4, \ldots \ldots$
iii)
Verify that each of the following is an AP, and then write its next three terms.
$\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, \ldots$
iv)
Verify that each of the following is an AP, and then write its next three terms.
a + b, (a + 1) + b, (a + 1) + (b + 1), ...
v)
Verify that each of the following is an AP, and then write its next three terms.
a, 2a + 1, 3a + 2, 4a + 3,...

i)
Here the given series is,
$0,\frac{1}{4},\frac{1}{2},\frac{3}{4}$
$\\ a\textsubscript{1} = 0, a\textsubscript{2}= \frac{1}{4} {{a}_{3}} = \frac{3}{4} \\ a\textsubscript{2} - a\textsubscript{1} = \frac{1}{4}-0=\frac{1}{4} \\$
$\\a_3-a_2=\frac{1}{4}\\a_4-a_3=\frac{1}{4}$
Here the common difference is the same. Hence given series is an AP
$\\a\textsubscript{5} = a\textsubscript{1}+4d (using a\textsubscript{n} = a\textsubscript{1 }+ (n-1).d)\\ \vspace{\baselineskip} a\textsubscript{6} = a\textsubscript{1}+ 5d \\ \vspace{\baselineskip} a\textsubscript{7} = a\textsubscript{1}+ 6d \\$
\begin{aligned} & =0+6\left( \frac{1}{4} \right) \\ & =\frac{3}{2} \ \end{aligned} \\
Next three terms are
$1,\frac{5}{4},\frac{3}{2} \\$
ii)
$\\Here the given series is, 5, \frac{14}{3},\frac{13}{3},4,..... \\ Here common difference is same. \\ Hence the given series is an AP\\ a\textsubscript{5} = {{a}_{1}} +4d (using a\textsubscript{n} = a\textsubscript{1} + (n-1).d)\\ a\textsubscript{6 }= {{a}_{1}} +5d \\ a\textsubscript{7} = {{a}_{1}} + 6d\\ \vspace{\baselineskip} Next terms are \frac{11}{3} , \frac{10}{3} ,3 \\$
iii)
$\\\text{Here the given series is,} \sqrt{3},2\sqrt{3},3\sqrt{3},.... \\ \text{Here common difference is same. Hence the given series is an AP}.\\a\textsubscript{4} = {{a}_{1}} +3d (using a\textsubscript{n} = {{a}_{1}} + (n-1).d)\\ =\sqrt{3}+3\sqrt{3}=4\sqrt{3} \\ a\textsubscript{5} = {{a}_{1}} + 4d\\ = \sqrt{3}+4\sqrt{3}=5\sqrt{3} \\ a\textsubscript{6} = {{a}_{1}} + 5d \\ = \sqrt{3}+5\sqrt{3}=6\sqrt{3} \\ Next terms are 4\sqrt{3},5\sqrt{3},6\sqrt{3} \\$
iv)
$\\\text{Here the given series is,} a + b, (a + 1) + b, (a + 1) + (b + 1) \ldots .\\ a\textsubscript{1} = a + b, a\textsubscript{2}= (a + 1) + b, a\textsubscript{3}= (a + 1) + (b + 1)\\ a\textsubscript{2} - a\textsubscript{1} = (a + 1) + b - a - b \\ = 1\\ a\textsubscript{3} - a\textsubscript{2} = a + 1 + b + 1 - a - 1 - b\\ = 1 \\ \text{Here common difference is same hence given series is an AP}\\ a\textsubscript{4} = a\textsubscript{1}+3d (using a\textsubscript{n} = a\textsubscript{1} + (n-1).d) \\ =(a + b) + 3(1)\\ (a + 1) + (b + 1) + 1\\ a\textsubscript{5} = a + 4d\\ =(a + b) + 4(1)\\ (a + 1) + (b + 1) + 1 + 1\\ a\textsubscript{6} = a + 5d\\ = (a + b) + 5(1)\\ (a + 1) + (b + 1) + 1 + 1 + 1\\ \text{Next three terms are:}\\ (a + 1) + (b + 1) + 1, (a + 1) + (b + 1) + 1 + 1, (a + 1) + (b + 1) + 1 + 1 + 1.\\$
v)
$\\\text{Here the given series is,} a, 2a + 1, 3a + 2, 4a + 3 \ldots \ldots \\ a\textsubscript{1} = a , a\textsubscript{2}=2a+1 , a\textsubscript{3}=3a + 2\\ a\textsubscript{2} - a\textsubscript{1}= 2a + 1 - a \\ = a + 1\\ a\textsubscript{3} - a\textsubscript{2} = 3a + 2 - (2a + 1)\\ =a + 1\\ \text{Here common difference is same hence given series is an AP}.\\ a\textsubscript{5} = a\textsubscript{1}+4d (using a\textsubscript{n} = a\textsubscript{1}+ (n-1)d) \\ =a + 4(a + 1)\\ = a + 4a + 4\\ = 5a + 4\\ a\textsubscript{6} = a\textsubscript{1} + 5d \\ = a + 5(a + 1)\\ = a + 5a + 5\\ = 6a + 5\\ a\textsubscript{7} = a\textsubscript{1} + 6d \\ = a + 6(a + 1)\\ = a + 6a + 6\\ = 7a + 6\\ \text{Next three terms are: }5a + 4, 6a + 5, 7a + 6 \ldots \\$

Question:3

i)
Write the first three terms of the APs when a and d are as given below:
$\mathrm{a}=\frac{1}{2}, \mathrm{~d}=-\frac{1}{6}$
ii)
Write the first three terms of the APs when a and d are as given below:
a = – 5, d = – 3
iii)
Write the first three terms of the APs when a and d are as given below:
$\mathrm{a}=\sqrt{2}, \mathrm{~d}=\frac{1}{\sqrt{2}}$

i)
\\Here a = \frac{1}{2} , d = -\frac{1}{6} \\ {{a}_{1}}=a=\frac{1}{2} \\ {{a}_{2}}=a+d (Using a\textsubscript{n} = a + (n-1).d)\\ a\textsubscript{3} = a + 2d\\ \begin{aligned} & =\frac{1}{2}+2\left( \frac{-1}{6} \right) \ & =\frac{1}{2}-\frac{2}{6} \ \end{aligned} \\ =\frac{3-2}{6}=\frac{1}{6} \\ Hence, first there terms are: \frac{1}{2},\frac{1}{3},\frac{1}{6},..... \\
ii)
$\\Here a = - 5, d = -3\\ a\textsubscript{1} = a = - 5\\ a\textsubscript{2}=a+d (Using a\textsubscript{n} = a + (n-1).d)\\ = - 5 - 3 = - 8\\ a\textsubscript{3} = a + 2d\\ = -5 + 2(-3)\\ = -5 - 6 = - 11\\ Here, first three terms are: -5, -8, -11.\\$
iii)
\\Here a = \sqrt{2},d=\frac{1}{\sqrt{2}} \\ a\textsubscript{2}=a+d (using a\textsubscript{n} = a + (n-1)d)\\ =\sqrt{2}+\frac{1}{\sqrt{2}} \\ =\frac{2+1}{\sqrt{2}}=\frac{3}{\sqrt{2}} \\ a\textsubscript{3} = a + 2d\\ =\sqrt{2}+2\left( \frac{1}{\sqrt{2}} \right) \\ =\sqrt{2}+\frac{2}{\sqrt{2}} \\ \begin{aligned} & =\frac{2+2}{\sqrt{2}} \ & =\frac{4}{\sqrt{2}} \ \end{aligned} \\ Here, first three terms are : \sqrt{2},\frac{3}{\sqrt{2}},\frac{4}{\sqrt{2}} \\

Question:4

Find a, b and c such that the following numbers are in AP: a, 7, b, 23, c.

Given AP is a, 7, b, 23, c
We know that if series are in AP then
$a\textsubscript{1}= a, a\textsubscript{2}= 7, a\textsubscript{3}= b, a\textsubscript{4}=23, a\textsubscript{5}= c \\$
(We know that the common difference of an A.P is Equal)
$\\a\textsubscript{2} - a\textsubscript{1} = a\textsubscript{3} - a\textsubscript{2 }= a\textsubscript{4} - a\textsubscript{3} = a\textsubscript{5} - a\textsubscript{4}\\ \therefore 7 - a = b - 7 = 23 - b = c - 23\\ Taking 2\textsuperscript{nd} and 3\textsuperscript{rd} term\\ b - 7 = 23 - b\\ 2b = 30\\ b=\frac{30}{2}=15 \\ Taking 1\textsuperscript{st} and 2\textsuperscript{nd} term\\ 7 - a = b - 7\\ 7-a=15-7 ( \because b = 15)\\ 7 - 8 = a\\ -1 = a\\ Taking 1\textsuperscript{st} and fourth term\\ 7 - a = c - 23\\ 7-(-1)=c-23 (\because a = -1)\\ 7 + 1 = c - 23\\ 8 + 23 = 6\\ 31 = c \\$
Hence a = - 1, b = 15, c = 31

Question:5

Determine the AP whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20.

$\\\text{Given } a\textsubscript{5} = 19\\ a + 4d = 19 \ldots (1) (\text{using} a\textsubscript{n} = a + (n-1)d)\\ , a\textsubscript{13} - a\textsubscript{8} = 20\\ a + 12d - a - 7d = 20\\ 5d = 20\\ d=\frac{20}{5} \\ d = 4\\ \text{Put d }= 4 in (1) we get\\ a + 4(4) = 19\\ a = 19 - 16\\ a = 3\\ a\textsubscript{1} = a = 3\\ a\textsubscript{2} = a + d = 3 + 4 = 7\\ a\textsubscript{3}= a + 2d = 3 + 2(4) = 3 + 8 = 11\\ a\textsubscript{4} = a + 3d = 3 + 3(4) = 3 + 12 = 15\\ Hence the AP is: 3, 7, 11, 15 \ldots \\$

Question:6

The 26th, 11thand the last term of an AP are 0, 3 and -1/5 , respectively. Find the common difference and the number of terms.

$\\Given a\textsubscript{26} = 0, a\textsubscript{11} = 3, a\textsubscript{n} = - \frac{1}{5} \\ a + 25d = 0 \ldots .(1) (using a\textsubscript{n} = a + (n-1)d)\\$
$\\a + 10d +3 \ldots .(2)(using a\textsubscript{n} = a + (n-1)d)\\ \\ 15d = - 3\\$
$\\Put d = -\frac{1}{5} in (1) we get\\ a - 5 = 0\\ a = 5\\ a\textsubscript{n }=\textsubscript{ } -\frac{1}{5} \\$
$We know that a\textsubscript{n} = a + (n - 1)d\\$
$\\\Rightarrow -\frac{1}{5}-5=\left( n-1 \right)\left( -\frac{1}{5} \right) \\ \Rightarrow \frac{-1-25}{5}=\left( n-1 \right)\left( -\frac{1}{5} \right) \\ \Rightarrow \frac{-26}{5}=\left( n-1 \right)\left( -\frac{1}{5} \right) \\ \Rightarrow \frac{-26}{5}\times \frac{-5}{1}=n-1 \\ \therefore 26 = n - 1\\ \Rightarrow n = 26 + 1\\ \Rightarrow n = 27 \\ ANS;- \\ d= -\frac{1}{5} , n = 27\\$

Question:7

The sum of the 5th and the 7th terms of an AP is 52 and the 10th term is 46. Find the AP.

$\\Given a\textsubscript{5} + a\textsubscript{7}=52 \\ a\textsubscript{5}=a+4d a\textsubscript{7}= a + 6d (using a\textsubscript{n} = a + (n-1).d) \\ Given a\textsubscript{10}= 46 \\ \therefore a+9d=46 \ldots .(1) \\ a + 4d + a + 6d = 52\\ 2a + 10d = 52\\ Dividing both side by 2\\ a + 5d = 26 \ldots (2)\\ a + 9d =\_ 46 \\ \\ -4d = - 20\\ d=\frac{-20}{-4} \\ d = 5 \\ Put d = 5 in equation (1) we get\\ a + 5 \times (5) = 26\\ a + 25 = 26\\ a = 26 - 25\\ a = 1\\ a\textsubscript{1} = a = 1\\ a\textsubscript{2} = a + d \\ = 1 + 5\\ = 6\\ a\textsubscript{3} = a + 2d \\ = 1 + 2 \times (5) \\ = 1 + 10 \\ = 11\\$

Hence the AP is 1, 6, 11....

Question:8

Find the 20th term of the AP whose 7th term is 24 less than the 11thterm, first term being 12.

Given a = 12
$\\a\textsubscript{7} = a\textsubscript{11} - 24\\ a+6d=a+10d-24 (using a\textsubscript{n} = a + (n-1).d)\\ a - a + 24 = 10d - 6d\\ 24 = 4d \\ d = 6\\$
We know that
$a\textsubscript{n} = a + (n - 1)d\\$
$\\Put n = 20, a\textsubscript{20} = a + (20 - 1)d\\ = 12 + (19) \times (6)\\ = 12 + 114\\ =126\\$
Hence
$20\textsuperscript{th} \text{term is 126}$

Question:9

If the 9th term of an AP is zero, prove that its 29th term is twice its 19th term.

Let first term = a
common difference = d
Given:- $a\textsubscript{9} = 0\\$
$\\\Rightarrow \therefore a + 8d = 0 (using a\textsubscript{n} = a + (n-1).d)\\ \Rightarrow \therefore a=-8d \ldots (1) \\ \text{To prove: } 2a\textsubscript{19} = a\textsubscript{29}\\ a\textsubscript{19}=a+18d (using a\textsubscript{n} = a + (n-1)d)\\ = - 8d + 18d \{ a = - 8d \text{ from equation (1)} \} \\ = 10d \ldots (2)\\ a\textsubscript{29} = a + (29 - 1)d\\ = a + 28d\\ = - 8d + 28d \{ a = - 8d from equation (1) \} \\ = 20d \\ =2 \times 10d\\ \text{By equation (2)}\\ a\textsubscript{29 }= 2a\textsubscript{19} \\$
Hence proved

Question:10

Find whether 55 is a term of the AP: 7, 10, 13, ______or not. If yes, find which term it is.

Let 55 is the nth term of an AP
$\therefore a\textsubscript{n} = 55 \ldots (1)\\$
Given AP is 7, 10, 13........
Here a = 7, d = 10 - 7
d = 3
By equation (1) is $a\textsubscript{n} = 55\\$
$\\a + (n - 1) \times d = 55 \ldots (2) (using a\textsubscript{n} = a + (n-1)d)\\ 7 + (n - 1) \times 3 = 55\\ (n - 1) \times 3 = 55 - 7\\ (n - 1) \times 3 = 48\\ n - 1 = \frac{48}{3} \\ n - 1 = 16\\ n = 17\\$
Here n is a positive integer.
Hence 55 is 17th term of an AP

Question:11

Determine k so that $k^2+ 4k + 8, 2k^2 + 3k + 6, 3k^2 + 4k + 4$ are three consecutive terms of an AP.

Here the given AP is
$k\textsuperscript{2} + 4k + 8, 2k\textsuperscript{2} + 3k + 6, 3k\textsuperscript{2} + 4k + 4\\$
$a\textsubscript{1}= k\textsuperscript{2} + 4k + 8, a\textsubscript{2}=2k\textsuperscript{2}+3k+6, a\textsubscript{3}=3k\textsuperscript{2} + 4k + 4\\$
$\\a\textsubscript{2}- a\textsubscript{1} = 2k\textsuperscript{2} + 3k + 6 - (k\textsuperscript{2} + 4k+ 8)\\ = 2k\textsuperscript{2} + 3k + 6 - k\textsuperscript{2} - 4k - 8\\ = k\textsuperscript{2} - k - 2\\ a\textsubscript{3} - a\textsubscript{2} = 3k\textsuperscript{2} + 4k + 4 - (2k\textsuperscript{2}+3k + 6) \\ = 3k\textsuperscript{2} + 4k + 4 - 2k\textsuperscript{2} - 3k - 6\\ = k\textsuperscript{2} + k - 2\\$
Hence, it is given that these terms are an AP
So the common difference of the A.P
$\\a\textsubscript{2} - a\textsubscript{1} = a\textsubscript{3} - a\textsubscript{2}\\ k\textsuperscript{2} - k - 2 = k\textsuperscript{2} + k - 2\\ k\textsuperscript{2} - k - 2 - k\textsuperscript{2} - k + 2 = 0\\ -2k = 0\\ k = 0\\$

Question:12

Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.

Let the three parts are (a - d), a, (a + d)
As per the question -
$\\ a - d + a + a + d = 207\\ 3a = 207\\ a = \frac{207}{3}=69 \\ a = 69\\$
It is given that the product of the two similar parts is 4623
$\\ (a - d) \times a = 4623\\ (69 - d) \times 69 = 4632\\ \\ 69 - 67 = d\\ d = 2\\ \text{So the required terms are}\\ a - d = 69 - 2 = 67\\ a = 69\\ a + d = 69 + 2 = 71\\$
Required numbers are 67, 69, and 71

Question:13

The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.

Let the three angles are (a - d), a, (a + d) are in AP
According to question the greatest is twice the least
$\\(a + d) = 2(a - d)\\ a + d = 2a - 2d\\ 2a - a - 2d - d = 0\\ a - 3d = 0\\ a = 3d \ldots (1)\\$
We know that the sum of angles of a triangle is $180 ^{\circ}$
$\\a + a - d + a + d = 180\textsuperscript{0}\\ 3a = 180\textsuperscript{0}\\ a=\frac{{{180}^{0}}}{3}={{60}^{0}} \\ a = 60\textsuperscript{0}\\ Put\ a = 60\textsuperscript{0} in (1) \\ 60\textsuperscript{0} = 3d \\ So the angles are \\ a - d = 60\textsuperscript{0} - 20\textsuperscript{0} = 40\textsuperscript{0}\\ a = 60\textsuperscript{0}\\ a + d = 60\textsuperscript{0} + 20\textsuperscript{0} = 80\textsuperscript{0}\\$
Required angles
$40\textsuperscript{0 }, 60\textsuperscript{0 } ,80\textsuperscript{0}\\$

Question:14

If the nth terms of the two APs: 9, 7, 5, ... and 24, 21, 18,... are the same, find the value of n. Also find that term.

The given APs are
$9, 7, 5, \ldots \ and \ 24, 21, 18, \ \ \ldots$
In AP
$9, 7, 5, \ldots ..\\$
$\\ a\textsubscript{1} = a = 9 , a\textsubscript{2}=7\\ d\textsubscript{1} = a\textsubscript{2} - a\textsubscript{1} = 7 - 9 = - 2\\ a\textsubscript{n} = a + (n - 1) d\textsubscript{1 }\\ a\textsubscript{n} = 9 + (n - 1) (-2) \ldots (1)\\ \text{In AP 24, 21, 18} \ldots \\ b\textsubscript{1} =b = 24, b\textsubscript{2 }= 21\\ d\textsubscript{2} = b\textsubscript{2} - b\textsubscript{1} = 21 - 24 = - 3\\ b\textsubscript{n} = b + (n - 1)d\textsubscript{2}\\ b\textsubscript{n}=24+ (n - 1) (-3) \ldots .(2) \\$
Here the number of terms n is equal in both case
According to question
$a\textsubscript{n} = b\textsubscript{n}\\$
$\\9 + (n - 1) (-2) = 24 + (n - 1) (-3)\\ 9 - 2n + 2 = 24 - 3n + 3\\ 9 - 2n + 2 - 24 + 3n - 3 = 0\\ n - 16 = 0\\ n = 16\\ a\textsubscript{16}=a+(16-1)d (using \ a\textsubscript{n} = a + (n-1)d)\\ = 9 + 15(-2)\\ = 9 - 30 = - 21\\ a\textsubscript{16} = - 21\\ n\textsuperscript{th} \text{term is - 21 and the value of n is 16}\\$

Question:15

If sum of the 3rd and the 8th terms of an AP is 7 and the sum of the 7thand the 14thterms is –3, find the 10th term.

It is given that the sum of $3\textsuperscript{rd} and 8\textsuperscript{th} term is 7$.

$\\a\textsubscript{3} = a + (3 - 1)d (using a\textsubscript{n} = a + (n-1)d)\\ a\textsubscript{3} = a + 2d\\ a\textsubscript{8} = a + (8 - 1)d (using a\textsubscript{n} = a + (n-1)d)\\ a\textsubscript{8} = a + 7d\\ a\textsubscript{3} + a\textsubscript{8 }= 7\\ a + 2d + a +7d = 7\\ 2a + 9d = 7 \ldots (1)\\ \text{Similarly the sum of 7}\textsuperscript{th} and 14\textsuperscript{th} term is - 3\\ a\textsubscript{7} + a\textsubscript{14} = - 3\\ a + (7 - 1).d + a + (14 - 1).d = - 3 (using a\textsubscript{n} = a + (n-1)d)\\ a + 6d + a + 13d = - 3\\ 2a + 19d = - 3 \ldots (2)\\$
$\\Solve (1) and (2) 2a + 19d = - 3\\ 2a + 9d = 7\\ 10d = - 10\\ d = - 1\\ Put d = - 1 in (1)\\ \Rightarrow 2a + 9(-1) = 7 \\ \Rightarrow 2a = 7 + 9 \\ \Rightarrow a=\frac{16}{2}=8 \\ a\textsubscript{10} = (a + (10 - 1)d) \\ \Rightarrow a\textsubscript{10} = 8 + 9(-1) \\ \Rightarrow a\textsubscript{10} = 8 - 9\\ \Rightarrow a\textsubscript{10} = -1\\$

Question:16

Find the 12thterm from the end of the AP: -2, -4, -6,..., -100.

The given AP : is $- 2, - 4, -6, \ldots \ldots . , - 100\\$
AP from end is $\dpi{100} - 100, - 98 \ldots \ldots \ldots .. 4, -6, -2\\$
$\dpi{100} \\ a\textsubscript{1}= a = - 100\\ d = a\textsubscript{2} - a\textsubscript{1} = - 98 - (-100) = - 98 + 100 = 2\\ a\textsubscript{12} = a + (12 - 1).d (using a\textsubscript{n} = a + (n-1) \times d)\\ a\textsubscript{12} = - 100 + 11 \times (2)\\ a\textsubscript{12} = - 100 + 22\\ a\textsubscript{12} = - 78\\$

Question:17

$\\\text{The given AP:- 53, 48, 43,} \ldots \ldots\\ a = 53\\ d = 48 - 43 = - 5\\ \text{We know that the negative term is less than 0}.\\ \text{Let nth term is negative term}.\\ a\textsubscript{n}< 0\\ a + (n - 1) \times d< 0 (using a\textsubscript{n} = a + (n-1)d)\\ 53 + (n - 1) \times (- 5) < 0\\ (n - 1) \times (-5) < - 53\\ -5n + 5 < -53\\ -5n < -58\\ 5n >58\\ \text{ Hence, }12\textsuperscript{th} term is the first negative term.\\$

Question:18

How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?

$\\The numbers between 10 and 300 which when divided by 4 leave remainder 3 is-\\ 11, 15, 19, 23, \ldots ., 299\\ a\textsubscript{1}=11, a\textsubscript{2}=15, a\textsubscript{3}=19\\ a\textsubscript{2} - a\textsubscript{1} = 15 - 11 = 4\\ a\textsubscript{3} - a\textsubscript{2} = 19 - 15 = 4\\ it is a\textsubscript{n} AP with a\textsubscript{1}=a = 11, d = 4\\ Let there are n terms in this AP\\ a\textsubscript{n} = 299\\ a + (n - 1) \times d = 299 (using a\textsubscript{n} = a + (n-1)d)\\ 11 + (n - 1) \times 4 = 299\\ (n - 1) \times 4 = 299 - 11\\ (n - 1) \times 4 = 288\\ (n-1)=\frac{288}{4} \\ (n - 1) = 72\\ n = 72 + 1 \\ n = 73\\$

Question:19

Find the sum of the two middlemost terms of the AP:
$-\frac{4}{3},-1,-\frac{2}{3}, \ldots, 4 \frac{1}{3}$

Here the given AP is
$-\frac{4}{3},-1,-\frac{2}{3},....,4\frac{1}{3}$
$\\ In this AP: {{a}_{1}} = a=-\frac{4}{3} \\ d={{a}_{2}}-{{a}_{1}}=-1-\left( -\frac{4}{3} \right) \Rightarrow =-1+\frac{4}{3}=\frac{-3+4}{3}=\frac{1}{3} \\$
Let there are n terms in AP
The nth term of AP is
$\\4\frac{1}{3} .\\ {{a}_{n}}=4\frac{1}{3} \\ a+(n-1).d=\frac{13}{3} (Using a\textsubscript{n} = a + (n-1)d)\\$
$\\-\frac{4}{3}+(n-1)\times \left( \frac{1}{3} \right)=\frac{13}{3} \\ \left( n-1 \right)\times \left( \frac{1}{3} \right)=\frac{13+4}{3} \\ n - 1 = 17\\ n = 18\\$
Total terms in this AP is 18 and the middlemost terms and 9thand 10th.
$\\a\textsubscript{9} + a\textsubscript{10} = a + (a - 1)d + a + (10 - 1)d (using a\textsubscript{n} = a + (n-1)d)\\ = 2a + 17d\\$
$\\ =2\left( -\frac{4}{3} \right)+17\left( \frac{1}{3} \right)\\=\frac{-8}{3}+\frac{17}{3} \\ {{a}_{9}}+{{a}_{10}}=\frac{-8+17}{3}=\frac{9}{3}=3 \\ a\textsubscript{9} + a\textsubscript{10} = 3\\$

Question:20

The first term of an AP is -5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.

$\\Given that a = - 5, a\textsubscript{n} = 45, S\textsubscript{n} = 120\\ a\textsubscript{n}= 45\\ a + (n - 1) \times d = 45 (using a\textsubscript{n} = a + (n-1).d)\\ -5 + (n-1) \times d = 45\\ (n - 1) \times d = 50 \ldots (1)\\ S\textsubscript{n} = 120\\ {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] = 120\\ \frac{n}{2}\left[ -10+50 \right]=120 Using (1)\\ n \times (40) = 240\\ n = 24/4 = 6\\ Number of terms (n)= 6\\ Put n = 6 in (1)\\ (6 - 1) \times d = 50\\ 5d = 50\\ d = 10\\ Common difference = 10\\$

Question:21

i)Find the sum:1 + (-2) + (-5) + (-8) + ... + (-236)
ii)Find the sum:$4-\frac{1}{n}+4-\frac{2}{n}+4-\frac{3}{n}+\ldots$ up to n terms
iii) Find the sum:$\frac{a-b}{a+b}+\frac{3 a-2 b}{a+b}+\frac{5 a-3 b}{a+b}+.$ to 11 terms

i)
The given series is
$1 + (-2) + (-5) + (-8) + \ldots + (-236)\\$
Here
$a\textsubscript{1}=a = 1 , a\textsubscript{2}= -2\\$
$\\d = a\textsubscript{2} - a\textsubscript{1} = - 2 - 1 = - 3\\ a\textsubscript{n} = - 236\\ a + (n - 1) \times d = - 236 ( a\textsubscript{n} = a + (n - 1)d)\\ 1 + (n - 1) \times (-3) = - 236\\ (n - 1) \times (-3) = - 236 - 1\\ n - 1 = 79\\ n = 80\\$
The formula of the sum is :
${{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\$
$\mathrm{S}_{\mathrm{n}}=\frac{80}{2}[2 \times(1)+(80-1) \times(-3)]$
$\\= 40 [2 + 79 \times (-3)]\\ = 40 [2 - 237]\\ = 40 \times - 235 \\ S\textsubscript{n} = - 9400\\$
ii)
$\\The given series is \left[ 4-\frac{1}{n} \right]+\left[ 4-\frac{2}{n} \right]+\left[ 4-\frac{3}{n} \right]+.... `\\ First\,term(a)=\left[ 4-\frac{1}{n} \right] \\ d=\left[ 4-\frac{2}{n} \right]-\left[ 4-\frac{1}{n} \right] \\ =\frac{-2}{n}+\frac{1}{n} \\ =\frac{-1}{n} \ \\ \because {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\$
$\\S_n=\frac{n}{2}[2(4-\frac{1}{n})+\frac{1}{n}-1]\\\\=\frac{7n-1}{2}$
iii)
$\\\text{The given series is} \frac{a-b}{a+b}+\frac{3a-2b}{a+b}+\frac{5a-3b}{a+b}+.... \\ Here \ {a}_{1}=\frac{a-b}{a+b} {{a}_{2}}=\frac{3a-2b}{a+b} \\ d= a\textsubscript{2 - } a\textsubscript{1} \text{here d = common difference }\\ \text{Given that n = 11} \\ \\$
\begin{aligned} \mathrm{S}_{11} &=\frac{11}{2}\left[2 \mathrm{a}_{1}+(11-1) \mathrm{d}\right] &\left(\because \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(2 \mathrm{a}_{1}+(\mathrm{n}-1) \mathrm{d}\right)\right.\\ &=\frac{11}{2}\left[\frac{2(\mathrm{a}-\mathrm{b})}{\mathrm{a}+\mathrm{b}}+\frac{10(2 \mathrm{a}-\mathrm{b})}{\mathrm{a}+\mathrm{b}}\right] \\ &=\frac{11}{2}\left[\frac{2 \mathrm{a}-2 \mathrm{~b}+20 \mathrm{a}-10 \mathrm{~b}}{\mathrm{a}+\mathrm{b}}\right] \\ &=\frac{11}{2}\left[\frac{22 \mathrm{a}-12 \mathrm{~b}}{\mathrm{a}+\mathrm{b}}\right] \\ &=\frac{11}{2}\left[\frac{2(11 \mathrm{a}-6 \mathrm{~b})}{\mathrm{a}+\mathrm{b}}\right] \\ & \mathrm{S}_{11}=\frac{11(11 \mathrm{a}-6 \mathrm{~b})}{(\mathrm{a}+\mathrm{b})} \end{aligned}

Question:22

Which term of the AP: -2, -7, -12,... will be -77? Find the sum of this AP upto the term -77.

$\text{The given AP is -2, -7, -12 } \ldots \\ \text{Here first term (a)} = - 2\\ \text{Common difference (d)} = - 7 - (-2) = -7 + 2 = - 5\\ Let \ n\textsuperscript{th} term is - 77\\ Then a\textsubscript{n} = - 77\\ a + (n - 1).d = - 77\\ -2 + (n - 1).(-5) = - 77\\ (n - 1).(-5) = - 77 + 2\\ (n - 1) = \frac{-75}{-5}=15 \\ n = 15+1\\ n = 16\\ \text{Hence} 16\textsuperscript{th} term is -77\\\text{ Sum of 16 terms} \\$
\begin{aligned} \mathrm{S}_{16} &=\frac{16}{2}[2(-2)+(16-1)(-5)] \quad\left(\text { Using } \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]\right) \\ &=8[-4+15(-5)] \\ &=8[-4-75] \\ \mathrm{S}_{16} &=8 \times-79 \\ \mathrm{~S}_{16}=&-632 \end{aligned}

Question:23

If $a_n = 3 - 4n,$ $\text{show that} a\textsubscript{1}, a\textsubscript{2}, a\textsubscript{3} ,...$ form an AP. Also find $S\textsubscript{20}$

Here it is given that

$a\textsubscript{n} = 3 - 4n\\$

Put n =1 then

$a\textsubscript{1} = 3 - 4(1)\\$

$a\textsubscript{1} = 3 - 4= - 1\\$

Put n =2 then

$a\textsubscript{2} = 3 - 4(2)\\$

$a\textsubscript{2} = 3 \textbf{-} 8 = \textbf{- }5\\$

Put n =3 then

$a\textsubscript{3} = 3 - 4(3)\\$

$a\textsubscript{3} = 3 - 12 = - 9\\$

$\text{Then A.P is -1, -5, -9} \ldots \ldots \\ a\textsubscript{2} - a\textsubscript{1}= - 5 - (-1) = - 5 + 1 = - 4\\ a\textsubscript{3} - a\textsubscript{2} = - 9 - (-5) = - 9 + 5 = - 4\\ a\textsubscript{2} - a\textsubscript{1} = a\textsubscript{3} - a\textsubscript{2}\\ \text{ Hence}, a\textsubscript{1}, a\textsubscript{2}, a\textsubscript{3} \ldots \text{ forms an AP with } a\textsubscript{1 }= a = - 1 , d = -4\\ \text{ Here d= common difference and first term is a} = a\textsubscript{1}= - 1\\S_{2}=\frac{20}{2}\left[ 2\left( -1 \right)+\left( 20-1 \right)\left( -4 \right) \right] \left( Using \ S_{n}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \right) \\ = 10 [- 2 + 19(-4)]\\ = 10[-2 - 76]\\ S\textsubscript{20} = - 780\\$

Question:24

$\\\text{It is given that S\textsubscript{n}} = n (4n + 1) \ldots . (1)\\ We know that \\ Put n=1 in equation (1) here S\textsubscript{1} =first term( a\textsubscript{1} ) \\ S\textsubscript{1} = 1(4(1) + 1)\\ S\textsubscript{1} = 5\\ S\textsubscript{1} = a\textsubscript{1} = 5\\ S\textsubscript{2} Sum of the first 2 terms of A.P\\ Put n = 2 in equation (1) \\ S\textsubscript{2} = 2(4(2) + 1)\\ S\textsubscript{2} = 2(9) = 18 \\ a\textsubscript{1}+a\textsubscript{2} =18 a\textsubscript{n}=a\textsubscript{1}+ (n - 1).d\\ a\textsubscript{1 }+ a\textsubscript{1} + d = 18\\ 2 a\textsubscript{1} + d = 18\\ 2(5) + d = 18 \\ d = 18 - 10\\ d = 8\\ Terms of an AP are a\textsubscript{1}, a\textsubscript{1} + d, a\textsubscript{1} + 2d \ldots ..\\ a\textsubscript{1} = 5 \\ a\textsubscript{1}+ d = 5 + 8 = 13 \\ a\textsubscript{1}+ 2d = 5 + 2(8)\\ = 5 + 16 = 21 \\ So the required AP is 5, 13, 21, \ldots \ldots \\$

Question:25

In an AP, if $S\textsubscript{n} = 3n\textsuperscript{2} + 5n and a\textsubscript{k} = 164,$ find the value of k

$\\\text{Here it is given that} S\textsubscript{n} = 3n\textsuperscript{2} + 5n \ldots \ldots (1)\\ a\textsubscript{k} = 164\\ We know that S\textsubscript{1} = a, S\textsubscript{2} = a + (a + d)\\ Put n=1 in equation (1)\\ S\textsubscript{1} = 3(1)\textsuperscript{2} + 5(1) = 8\\ Put n=2 in equation (1)\\ S\textsubscript{2} = 3(2)\textsuperscript{2} + 5(2)\\ S\textsubscript{2} = 12 + 10 = 22\\ a = 8 \\ a + a + d = 22\\ 2a + d = 22 \\ 2(8) + d = 22\\ d = 22 - 16\\ d = 6\\ The AP is a, a + d, a + 2d \ldots \ldots .\\ a = 8\\ a + d = 8 + 6 = 14\\ a + 2d = 8 + 6(2) = 8 + 12 = 20\\ 8, 14, 20 \ldots \\ a\textsubscript{k} = 164\\ a + (k - 1). d = 164\\ 8 + (k - 1). 6 = 164\\ (k - 1).6 = 164 - 8\\ k = 26+1\\ k = 27\\$

Question:26

If $S\textsubscript{n}$ denotes the sum of first n terms of an AP, prove that $\\ S\textsubscript{12} = 3(S\textsubscript{8} - S\textsubscript{4})\\$

$\\\text{To prove}:- S\textsubscript{12} = 3(S\textsubscript{8} - S\textsubscript{4})\\ If \ S\textsubscript{n} \text{is the sum of n terms then first term is a and common difference is d} \\ S_{n}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \ldots \ldots .. (1)\\ \text{Put }n = 12 in equation (1) \\ S\textsubscript{12} = 6[2a + 11d]\\ S\textsubscript{12} = 12a + 66d \ldots \ldots \ldots \ldots . (1)\\ S\textsubscript{8} = 4[2a + 7d]\\ S\textsubscript{8} = 8a + 28d \ldots \ldots \ldots \ldots \ldots .(2)\\ S\textsubscript{4} = 2[2a + 3d]\\ S\textsubscript{4} = 4a + 6d \ldots \ldots \ldots \ldots \ldots ..(3)\\ RHS \\3(S\textsubscript{8} - S\textsubscript{4}) = \\ =3(8a + 28d - 4a - 6d) \text{By Equation (2) and (3)}\\ =3(4a + 22d)\\ = 12a + 66d By Equation (1)\\ = S\textsubscript{12}\\ \text{Hence } \textsubscript{12} = 3(S\textsubscript{8} - S\textsubscript{4})$

Question:27

Find the sum of first 17 terms of an AP whose 4th and 9th terms are -15 and -30 respectively.

$\\a\textsubscript{4 }= - 15, a\textsubscript{9} = - 30\\ a\textsubscript{4} = - 15 a\textsubscript{9} = - 30\\ a + 3d = - 15 \ldots (1) a + 8d = - 30 \ldots (2) (Using a\textsubscript{n} = a + (n - 1).d)\\ \text{from equation (1)} \& (2) \\$
\\-5d = 15\\ d = - 3\\ Put d = - 3 in (1)\\ a + 3(-3) = - 15\\ a = - 15 + 9\\ a = - 6\\ {{S}_{17}}=\frac{17}{2}\left[ 2\left( -6 \right)+\left( 17-1 \right)\left( -3 \right) \right] \\Using \ {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\ \begin{aligned} & =\frac{17}{2}\left[ -12-48 \right] \ & =\frac{17}{2}\left[ -60 \right] \ \end{aligned} \\ S\textsubscript{17} = - 510\\

Question:28

If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

$\\S\textsubscript{6} = 36, S\textsubscript{16} = 256\\ \left[ Using\,\,{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \right] \\ S\textsubscript{6} = 36 S\textsubscript{16} = 256\\ = \frac{6}{2}\left[ 2a+\left( 6-1 \right)d \right]=36 \frac{16}{2}\left[ 2a+\left( 16-1 \right)d \right]=256 \\ =3[2a + 5d] = 36 8[2a + 15d] = 256\\ =2a + 5d = 12 \ldots (1) \\2a + 15d = 32 \ldots (2)\\\text{from (1)and (2)}$
$\\ \_\_\_\_\_\_\_\_\_\_\_\_\\ 10d = 20\\ d = 2\\ Put \ d = 2\ in (1)\\ 2a + 5(2) = 12\\ 2a = 12 - 10\\ a = 1\\ ( \left[ Using\,\,{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \right] \\ Put n= 1\\ = 5 [2 + 18] \\ 5[20]\\ S\textsubscript{10} = 100\\$

Question:29

Find the sum of all the 11 terms of an AP whose middlemost term is 30.

$\\The middle term of an AP = \frac{n+1}{2} \\ Here n = 11 \\ Hence a\textsubscript{6} = 30\\ a + 5d = 30 \ldots (1) (Using a\textsubscript{n} = a + (n - 1)d)\\ {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] put n=11\\ =\frac{11}{2}\left[ 2a+10d \right] \\ =\frac{11}{2}\left[ 2\left( a+5d \right) \right] \\ = 11 [30] (Using (1))\\ S\textsubscript{11} = 330\\$

Question:30

Find the sum of last ten terms of the AP: 8, 10, 12, __________ , 126.

$\\ \text{The given AP is 8, 10, } \ldots .., 126\\ \text{AP from last is 126, 124, } \ldots ., 10, 8\\ a = 126\\ d = 124 - 126 = - 2\\$
Sum of 10 terms from last-
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$Put \ n=10$
\begin{aligned} \mathrm{S}_{10} &=\frac{10}{2}[2(126)+(10-1)(-2)] \\ &=5[252+9(-2)] \\ &=5[252-18] \\ \mathrm{S}_{10} &=5 \times 234=1170 \\ \mathrm{~S}_{10} &=1170 \end{aligned}

Question:31

Find the sum of first seven numbers which are multiples of 2 as well as of 9.

$\\\text{LCM of 2 and 9 is 18.\\ So the terms are 18, 36, } \ldots .., 126\\ \text{First term (a) }= 18\\$
$\text{Common difference (d)} = 36 - 18 = 18\\$
${{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
$\\ Put\ n=7\\$
$\\ =\frac{7}{2}\left[ 36+108 \right] \\ =\frac{7}{2}\left[ 144 \right] \\$
$\\S\textsubscript{7} = 7 \times 72 \\ S\textsubscript{7} = 504\\$

Question:32

How many terms of the AP: -15, -13, -11, _______ are needed to make the sum -55? Explain the reason for double answer.

Here the given AP is $-15, -13, -11 \ldots \\$
First term (a) = - 15
Common difference (d) = - 13 - (-15) = - 13 + 15 = 2
Let n terms have sum - 55 then
${{S}_{n}}=-55$
${{S}_{n}} =\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
$\\ n[-30 + 2n - 2] = - 55 \times 2\\ -32n + 2n\textsuperscript{2} + 110 = 0\\ 2n\textsuperscript{2} - 32n + 110 = 0\\ n\textsuperscript{2} - 16n + 55 = 0\\ n\textsuperscript{2} - 5n - 11n + 55 = 0\\ n(n - 5) - 11(n - 5) = 0\\ (n - 5) (n - 11) = 0\\ n = 5, 11\\$
There are two answers which are 5th when n=5 all terms are negative and 11th terms when n=11 the AP will contain positive numbers and negative terms so the resulting sum is - 55.

Question:33

The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is - 30 and the common difference is 8. Find n.

$\\ Given: a = 8, d\textsubscript{1} = 20 for first AP\\ b = - 30, d\textsubscript{2} = 8 for second AP\\ Given S\textsubscript{n} = S\textsubscript{2n}\\ \frac{\mathrm{n}}{2}\left[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}_{1}\right]=\frac{2 \mathrm{n}}{2}\left[2 \mathrm{~b}+(2 \mathrm{n}-1) \mathrm{d}_{2}\right] \\ \frac{\mathrm{n}}{2}[2(8)+(\mathrm{n}-1) 20]=\mathrm{n}[2(-30)+(2 \mathrm{n}-1)(8)] \\ \frac{\mathrm{n}}{2}[16+20 \mathrm{n}-20]=\mathrm{n}[-60+16 \mathrm{n}-8] \\ \frac{\mathrm{n}}{2}[20 \mathrm{n}-4]=\mathrm{n}[-68+16 \mathrm{n}] \\ \frac{2(10 \mathrm{n}-2)}{2}=-68+16 \mathrm{n} 10n - 16n = - 68 + 2\\ -6n = - 66\\ n = \frac{+66}{+6}=11 \\ n = 11\\$

Question:34

Kanika was given her pocket money on Jan 1st, 2008. She puts Re 1 on Day 1,Rs 2 on Day 2, Rs 3 on Day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money, and found that at the end of the month she still had Rs 100 with her. How much was her pocket money for the month?

According to the question she puts Rs. 1 on Day 1, 2 on Day 2 and so on.
$\\Hence 1, 2, 3, \ldots \ldots .. , 31\\ where a = 1\\ d = 2 - 1 = 1 \\ a\textsubscript{n} = 31\\ Total money she put, {{S}_{n}}=\frac{n}{2}\left[ a+{{a}_{n}} \right] \\$
$\\ \mathrm{S}_{\mathrm{n}}=\frac{31}{2}[1+31] \\\\ \mathrm{S}_{\mathrm{n}}=\frac{31}{2} \times 32 \Rightarrow 31 \times 16=496$
Total money = Total she put + Spent + left
= 496 + 204 + 100
Total = 800 Rs

Question:35

Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000?

$\\ \text{As per the question she saves 32 in first, 36 in second and as so on}\\ 32, 36, 40, \ldots \ldots \\ \text{Here, first term (a)} = 32\\ \text{Common difference (d)} = 36 - 32 = 4\\ Let in n month she saves Rs. 2000\\S_{n}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\ 2000=\frac{\mathrm{n}}{2}[2 \times 32+(\mathrm{n}-1) 4] 4000 = n[64 + 4n - 4]\\ 4000 = n[60 + 4n]\\ 4n\textsuperscript{2} + 60n - 4000 = 0\\\text{Divide by 4}\\ n\textsuperscript{2} + 15n - 1000 = 0\\ n\textsuperscript{2} + 40n - 25n - 1000 = 0\\ n(n + 40) - 25 (n + 40) = 0\\ (n + 40) (n - 25) = 0\\ n = - 40\text{(months cannot be negative, which is not possible)}, n = +25\\ Hence it takes 25 months to save Rs. 2000.\\$

## Question:1

The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.

$\\S\textsubscript{5} + S\textsubscript{7} = 167 \ldots (1)\\ S\textsubscript{10} = 235 \ldots (2)\\$
We know that
$\\S_{n}=\frac{n}{2}\left( 2a+\left( n-1 \right)d \right) \\ \text{Put n }= 5, {{S}_{5}}=\frac{5}{2}\left( 2a+\left( 5-1 \right)d \right) \\ {{S}_{5}}=\frac{5}{2}\left( 2a+4d \right) \\ Put n = 7, {{S}_{7}}=\frac{7}{2}\left( 2a+\left( 7-1 \right)d \right) \\ {{S}_{7}}=\frac{7}{2}\left( 2a+6d \right) \\ Put n = 10, {{S}_{10}}=\frac{10}{2}\left( 2a+\left( 10-1 \right)d \right) \\ S\textsubscript{10} = 5(2a + 9d)\\ \text{Now put the values of} S\textsubscript{5}, S\textsubscript{7} and S\textsubscript{7} \text{ in equation (1) and (2) we get}\\ \frac{5}{2}\left( 2a+4d \right)+\frac{7}{2}\left( 2a+6d \right)=167 \\ 5a + 10d + 7a + 21d = 167\\ 12a + 31d = 167 \ldots (3)\\ 5(2a + 9d) = 235\\ 2a+9d=\frac{235}{5} \\ 2a + 9d = 47 \ldots (4)\\ Solving equation (3) and (4)\\ 12a + 31d = 167 \ldots (3)\\ 12a + 54d = 282 \{ Multiply (4) by 6 \} \\$
$\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ -23d = - 115\\ d=\frac{-115}{-23} \\ d = 5\\ Put d = 5 in (3) we get\\ 12a + 31(5) = 167 \\ 12a = 167 - 155 \\ 12a = 12\\ a=\frac{12}{12}=1 \\ = 10 [2 + 95]\\ S\textsubscript{20} = 970\\$

Question:2

i)Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
ii)Find the sum of those integers from 1 to 500 which are multiples of 2 as well as of 5 .
iii)Find the sum of those integers from 1 to 500 which are multiples of 2 or 5.

i)
$\\The numbers which are multiples of 2 as well as 5 i.e. multiple of 10.\\ 10, 20, 30 \ldots \ldots . 490.\\ Here first term (a) = 10\\ Common difference (d) = 20 - 10 = 10\\ a\textsubscript{n} = 490\\ a + (n - 1).d = 490\\ 10 + (n - 1)10 = 490\\ (n - 1)10 = 490 - 10\\ \left( n-1 \right)=\frac{480}{10} \\ n = 48 + 1\\ n = 49\\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]=\frac{49}{2}[10+490]=\frac{49}{2}[500] \\ = 49[250] = 12250\\$
ii)
$\\The numbers which are multiples of 2 as well as 5 is the multiple of 10.\\ 10, 20, 30, \ldots .., 500\\ Here first term (a) = 10\\ Common difference (d) = 20 - 10 = 10\\ a\textsubscript{n} = 500 \because a\textsubscript{n}=a + (n - 1)d \\ a + (n - 1)d = 500\\ 10 + (n - 1)10 = 500\\ (n - 1) 10 = 500 - 10\\ n-1=\frac{490}{10} \\ n = 49 + 1\\ n = 50\\ {{S}_{n}}=\frac{n}{2}\left[ a+{{a}_{n}} \right] put n =50\\ 25[510] = 12750\\$
iii)
$\\\text{The numbers which are multiples of 2 are 2, 4, 6, 8, } \ldots \ldots , 500 \\ \text{Here first term} (a\textsubscript{1}) = 2, \text{common difference} (d\textsubscript{1}) = 4 - 2, n_{1}=\frac{500}{2}=250 \\ a_{n_1} = 500\\ \mathrm{S}_{\mathrm{n}_{1}}=\frac{\mathrm{n}_{1}}{2}\left[\mathrm{a}_{1}+\mathrm{a}_{\mathrm{n}_{1}}\right] \\ \mathrm{S}_{\mathrm{n}_{1}}=\frac{250}{2}[2+500] \\ \mathrm{S}_{\mathrm{n}_{1}}=125[502] \\ \mathrm{S}_{\mathrm{n}_{1}}=62750\\ \text{ The numbers which are multiples of 5 are 5, 10, 15, } \ldots .., 500\\ Here a\textsubscript{2} = 5, d\textsubscript{2} = 10 - 5 = 5\\ a _{{{n}_{2}}} = 500\\ {{S}_{{{n}_{2}}}}=\frac{{{n}_{2}}}{2}\left[ {{a}_{2}}+{{a}_{{{n}_{2}}}} \right] \Rightarrow {{S}_{{{n}_{2}}}}=\frac{100}{2}\left[ 5+500 \right]=50\left( 505 \right)=25250 \\ The numbers which are multiples of 10 are 10, 20, 30, \ldots .., 50\\ Here a\textsubscript{3} = 10, d\textsubscript{3} = 20 - 10 = 10\\ {{n}_{3}}=\frac{500}{10}=50 \\$
$\\ \mathrm{S}_{\mathrm{n}_{3}}=\frac{\mathrm{n}_{3}}{2}\left[\mathrm{a}_{3}+\mathrm{a}_{\mathrm{n}_{3}}\right] \\\\ \mathrm{S}_{\mathrm{n}_{3}}=\frac{50}{2}[10+500]=25[510]=12750$
The sum of integers from 1 to 500 which are multiples of 2 or 5 = sum of integers which are multiples of 2 + sum of integer which are multiple of 5 - the sum of integer which is multiple of 10
(we subtract the sum of integers multiple of 10 because those integers are multiples of both 2 and 5 )
$=\mathrm{S}_{\mathrm{r}_{1}}+\mathrm{S}_{\mathrm{n}_{2}}-\mathrm{S}_{\mathrm{n}_{3}}=62750+25250-12750$
= 88000 - 12750 = 75250

Question:3

The eighth term of an AP is half its second term and the eleventh term exceeds
one-third of its fourth term by 1. Find the 15th term.

According to question

$\\ \\a\textsubscript{8 }= a\textsubscript{2}/2 \\ 2a\textsubscript{8} = a\textsubscript{2}\\ 2(a + 7d) = a + d (Using a\textsubscript{n} = a + (n - 1)d ) \\ 2a + 14d - a - d = 0\\ a + 13d = 0 \ldots (1)\\ 3a\textsubscript{11} = a\textsubscript{4} + 3\\ 3(a + 10d) = a + 3d + 3 (Using a\textsubscript{n} = a + (n - 1)d ) \\ 3a + 30d - a - 3d = 3\\ 2a + 27d = 3 \ldots (2)\\ Apply: eq.(2) - 2 \times eq.(1) we get\\ 2a + 27d - 2a - 26d = 3\\ d = 3\\ Put d = 3 in (1) we get\\ a +13(3) = 0\\ a = - 39\\ Then a\textsubscript{15} = a + 14d (Using a\textsubscript{n} = a + (n - 1)d ) \\ = - 39 + 14(3)\\ = -39 + 42 \\ = 3\\$

Question:4

An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.

Total terms = 37, middle term =
$\frac{37+1}{2}=19 \\$
Three middle most terms, 18th , 19thand 20th, given
$a\textsubscript{18} + a\textsubscript{19} + a\textsubscript{20} = 225 \\$
\\a + 17d + a + 18d + a + 19d = 225 ( Using a\textsubscript{n} = a + (n - 1)d ) \\ 3a + 54d = 225\\ dividing by 3 we get\\ a + 18d = 75\\ a = 75 - 18d \ldots (1)\\ According to question\\ a\textsubscript{35} + a\textsubscript{36} + a\textsubscript{37} = 429 ( Using a\textsubscript{n} = a + (n - 1)d ) \\ a + 34d + a+ 35d + a + 36d = 429\\ 3a + 105d = 429\\ Dividing by three we get\\ a + 35d = 143\\ Put value from equation (1) we get\\ 75 - 18d + 35d = 143, \\ 17d = 143 - 75\\ d = 68, \\ \begin{aligned} & d=\frac{68}{17} \ & d=4 \ \end{aligned} \\ Put d = 4 in (1) we get\\ a = 75 - 72 = 3\\ a + d = 3 + 4 = 7\\ a + 2d = 3 + 2(4) = 11\\ Required AP is 3, 7, 11, \ldots \ldots .\\

Question:5

i)Find the sum of the integers between 100 and 200 that are divisible by 9
ii)Find the sum of the integers between 100 and 200 that are not divisible by 9

i)
Numbers between 100 and 200 which is divisible by 9 are 108, 117, 126, ... 198.
Here a = 108, d = 117- 108 = 9
$\\ a\textsubscript{n} = 198\\ a + (n - 1) \times d = 198\\ 108 + (n - 1) \times 9 = 198\\ (n - 1) \times 9 = 198 - 108\\ \left( n-1 \right)=\frac{90}{9} \\$
\\n = 10 + 1 = 11 \\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\ \text { Put } \mathrm{n}=11 \\ \begin{aligned} \mathrm{S}_{11} &=\frac{11}{2}[2(108)+(11-1) 9] \\ &=\frac{11}{2}[216+90] \\ &=\frac{11}{2} \times 306 \\ &=11 \times 153 \\ \mathrm{~S}_{11} &=1683 \end{aligned}
ii)
Numbers between 100 and 200 are 101, 102, 103, .. , 199
Here a = 101, d = 102 - 101 = 1
$\\a\textsubscript{n} = 199\\ a + (n - 1).d = 199 ( Using a\textsubscript{n} = a + (n - 1)d )\\ 101 + (n - 1)1 = 199\\ (n - 1) = 199 - 101\\ n = 98 + 1\\ n = 99\\ {{S}_{99}}=\frac{99}{2}\left[ 2\times 101+\left( 99-1 \right)\left( 1 \right) \right] \\$
\begin{aligned} & =\frac{99}{2}\left[ 202+98 \right] \ & =\frac{99}{2}\left[ 300 \right] \ \end{aligned} \\
$\\=99[150]\\ = 14850\\$
Numbers between 100 and 200 which is divisible by 9 are 108, 117, 126, ...198
Here b = 108,
$d\textsubscript{1} = 117- 108 = 9\\$
$\\ b\textsubscript{N} = 198\\ b + (N - 1) \times d\textsubscript{1 } = 198\\ 108 + (N - 1) \times 9 = 198\\ (N - 1) \times 9 = 198 - 108\\ \left( N-1 \right)=\frac{90}{9} \\$
\\N= 10 + 1 = 11\\ \mathrm{S}_{\mathrm{N}}=\frac{\mathrm{N}}{2}\left[2 \mathrm{~b}+(\mathrm{N}-1) \mathrm{d}_{1}\right] \\ \text { Put } \mathrm{N}=11 \\ \begin{aligned} \mathrm{S}_{11} &=\frac{11}{2}[2(108)+(11-1) 9] \\ &=\frac{11}{2}[216+90] \\ &=\frac{11}{2} \times 306 \\ &=11 \times 153 \\ \mathrm{~S}_{11} &=1683 \end{aligned}
the sum of the integers between 100 and 200 that is not divisible by 9
=Sum of all numbers - the sum of numbers which is divisible by 9
$\\ = 14850 - 1683 \\ = 13167\\$

Question:6

The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5thterm to the 21stterm, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.

According to question
$\\\frac{{{a}_{11}}}{{{a}_{18}}}=\frac{2}{3} \Rightarrow \frac{a+10d}{a+17d}=\frac{2}{3} (Using a\textsubscript{n} = a + (n - 1).d )\\ \\ 3a + 30d = 2a + 34d\\ 3a - 2a + 30d - 34d = 0\\ a - 4d = 0 a = 4d \ldots (1)\\$
$\\Ratio of 5\textsuperscript{th} and 21th term is \\ \frac{{{a}_{5}}}{{{a}_{21}}}=\frac{a+4d}{a+20d} \ldots (2)\\ Put a = 4d in (2) we get =\frac{4d+4d}{4d+20d}=\frac{8d}{24d} \\$
$\ \frac{\mathrm{a}_{5}}{\mathrm{a}_{21}}=\frac{1}{3} \\Ratio of \mathrm{S}_{5} to \mathrm{S}_{21} is \\\frac{\mathrm{S}_{5}}{\mathrm{~S}_{21}}=\frac{\frac{5}{2}[2 \mathrm{a}+4 \mathrm{~d}]}{\frac{21}{2}[2 \mathrm{a}+20 \mathrm{~d}]} \quad\left[\because \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]\right] \\\frac{\mathrm{S}_{5}}{\mathrm{~S}_{21}}=\frac{5[2 \mathrm{a}+4 \mathrm{~d}]}{21[2 \mathrm{a}+20 \mathrm{~d}]} \\Put a=4 d in equation (3) we get \\\frac{\mathrm{S}_{5}}{\mathrm{~S}_{21}}=\frac{5[2(4 \mathrm{~d})+4 \mathrm{~d}]}{21[2(4 \mathrm{~d})+20 \mathrm{~d}]}=\frac{5[8 \mathrm{~d}+4 \mathrm{~d}]}{21[8 \mathrm{~d}+20 \mathrm{~d}]}=\frac{5[12 \mathrm{~d}]}{21[28 \mathrm{~d}]}=\frac{60 \mathrm{~d}}{588 \mathrm{~d}} \\\frac{\mathrm{S}_{5}}{\mathrm{~S}_{21}}=\frac{5}{49}$

Question:7

Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to
$\frac{(a+c)(b+c-2 a)}{2(b-a)}$

Here
$a\textsubscript{1} = a, a\textsubscript{2} = b\\$
$\\ d = a\textsubscript{2} - a\textsubscript{1}\\ = b - a \\ a\textsubscript{n} = c \: \: \: \: \: \: \: \: \: \: \: \: \because a\textsubscript{n} =a + (n - 1)d \\ a\textsubscript{1} + (n - 1).d = c\: \: \: \: \: Put a\textsubscript{1} = a , d= b - a \\ (n - 1) (b - a) = c - a$
$\\ \mathrm{n}-1=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{b}-\mathrm{a}} \\ \mathrm{n}=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{b}-\mathrm{a}}+1 \\ \mathrm{n}=\frac{\mathrm{c}-\mathrm{a}+\mathrm{b}-\mathrm{a}}{\mathrm{b}-\mathrm{a}} \\ \mathrm{n}=\frac{\mathrm{c}+\mathrm{b}-2 \mathrm{a}}{\mathrm{b}-\mathrm{a}} \\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2 \mathrm{a}_{1}+(\mathrm{n}-1) \mathrm{d}\right] \\ =\frac{\mathrm{c}+\mathrm{b}-2 \mathrm{a}}{2(\mathrm{~b}-\mathrm{a})}[\mathrm{a}+\mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\ =\frac{(\mathrm{c}+\mathrm{b}-2 \mathrm{a})}{2(\mathrm{~b}-\mathrm{a})}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right] \quad\left[\because \mathrm{a}+(\mathrm{n}-1) \mathrm{d}=\mathrm{a}_{\mathrm{n}}\right] \\ \frac{(\mathrm{a}+\mathrm{c})(\mathrm{b}+\mathrm{c}-2 \mathrm{a})}{2(\mathrm{~b}-\mathrm{a})} \quad \quad\left[\because \mathrm{a}_{\mathrm{n}}=\mathrm{c}\right]$
Hence proved.

Question:8

Solve the equation- 4 + (-1) + 2 +...+ x = 437

$\\Given series :\\ -4 + (-1) + 2 + \ldots + x = 437\\ Here a = - 4\\ d = - 1 - (-4)\\ -1 + 4 = + 3\\ S\textsubscript{n} = 437 (given)\\ We know that {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\437=\frac{n}{2}[2(-4)+(n-1) 3]\\ 874 = n[- 8 + 3n - 3]\\ 874 = 3n\textsuperscript{2} - 11n\\ 3n\textsuperscript{2} - 11n - 874 = 0\\ 3n\textsuperscript{2} - 57n + 46n - 874 = 0\\ 3n (n - 19) + 46 (n - 19) = 0 \\ (n - 19) (3n + 46) = 0\\ n - 19 = 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2n + 16 = 0\\ n = 19 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: n = - 46/3\\ Hence n = 19 \\ a\textsubscript{n} = x \\ a + (n - 1)d = x (using a\textsubscript{n }= a + (n-1)d )\\ -4 + (19 - 1)3 = x\\ -4 + (18)3 = x\\ -4 + 54 = x\\ x = 50\\$

Question:9

Jaspal Singh repays his total loan of Rs 118000 by paying every month starting with the first installment of Rs 1000. If he increases the installment by Rs 100 every month, what amount will be paid by him in the 30thinstalment? What amount of loan does he still have to pay after the 30th installment?

Total loan = 118000
First installment (a) = 1000
Installment increased = 100
Hence d = 100
$30\textsuperscript{th} \text{installment} = a\textsubscript{30} = a + 29d \\$
= 1000 + 29(100)
= 1000 + 2900
= 3900
Amount of loan paid after 30 installments
= Total loan - 30th installments is total loan
$\\118000 - \frac{30}{2}\left[ 2\times 1000+\left( 30-1 \right)\times 100 \right] \\ 118000 - 15 [2000 + 2900]\\ 118000 - 15 [4900]\\ 118000 - 73500 = 44500\\$

Question:10

The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?

Let she fixes 13 flags on one side and 13 on other side of the middle flag.
distance covered for fixing first flag and return is 2m + 2m = 4m
distance covered for fixing second flag and return is 4m + 4m = 8m
distance covered for fixing third flag and return is 6m + 6m = 12m
Similarly for thirteen flags is
4m + 8m + 12m + …….
Here, a = 4
d = 8 - 4 = 4
n = 13
$\\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\ \mathrm{S}_{13}=\frac{13}{2}[2(4)+(13-1) 4] \\ \frac{13}{2}[56]=364$
Total distance covered = 364 × 2 = 728 m (for both side )
distance, only to carry the flag = 364.

## NCERT Exemplar Solutions Class 10 Maths Chapter 5 Important Topics:

• Formation of progression.
• Understanding of common difference and general term of the progression.
• How to calculate sum of N terms of the progression.
• NCERT Exemplar class 10 maths solutions chapter 5 discusses the methods to find out first-term and common difference if and nth term is given.
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## NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

 Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry & Its Equations Chapter 9 Circles Chapter 10 Constructions Chapter 11 Areas related to Circles Chapter 12 Surface Areas and Volumes Chapter 13 Statistics and Probability

## Features of NCERT Exemplar Class 10 Maths Solutions Chapter 5:

These Class 10 Maths NCERT exemplar chapter 5 solutions provide a detailed knowledge of arithmetic progression. Progression is defined as a sequence of numbers in which numbers progress in some given order. Generally, there are arithmetic progressions, geometric progressions, and harmonic progressions. The quantity and quality of questions covered in the exemplar are sufficient for a student to solve most of the Arithmetic Progressions based practice problems.

If a student diligently practices and understands the Class 10 Maths NCERT exemplar solutions chapter 5 Arithmetic Progressions, solving other books such as A textbook of Mathematics by Monica Kapoor, NCERT Class 10 Maths, RS Aggarwal Class 10 Maths, RD Sharma Class 10 Maths, et cetera won’t be of any issue.

Careers 360 is highly dedicated to provide the students with a strong and sound learning curve and for this reason, NCERT exemplar Class 10 Maths solutions chapter 5 pdf download feature is available to provide a seamless experience while studying NCERT exemplar Class 10 Maths chapter 5.

### Check Chapter-Wise Solutions of Book Questions

 Chapter No. Chapter Name Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

### Must, Read NCERT Solution Subject Wise

##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

### Also, Check NCERT Books and NCERT Syllabus here

1. Can two different Arithmetic Progressions have the same common difference?

Yes, they can have same common difference but they must have different first term then only these two progressions will be considered as different progression

2. Sum of any arithmetic progression can be zero. Comment.

Yes, the summation of AP can be zero but for that exactly one first term or common difference has to be negative and other has to be positive.

3. Is the knowledge of Arithmetic progression of class 10 sufficient for JEE Advanced exam?

Yes, the knowledge of AP given in class 10 is more than sufficient for solving any questions of AP at a higher level. If any student refers NCERT exemplar Class 10 Maths solutions chapter 5, it would be very useful for mathematics at higher levels

4. Is the chapter Arithmetic Progression important for Board examinations?

This chapter is important for Class 10 board exams as it comprises around 4-5% weightage of the whole paper.

5. What percentage of marks does this chapter of Arithmetic Progression holds in board examinations?

Generally, Arithmetics progression is considered among the important topics for board examinations and accounts for 8-10% marks of the total paper.

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

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1. Starting Point:

• Determine your starting point in Aligarh or the nearby area.

3. By Local Transport:

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Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9