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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.

NCERT Exemplar Class 10 Maths Solutions Chapter 5 Arithmetic Progression: Exercise- 5.1

Question:1

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

(A) 6

(B) 7

(C) 20

(D) 28

Answer:

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

Answer:

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

Answer:

\\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

The 11th term of the AP:
-5, \frac{-5}{2},0,\frac{5}{2}, $ \ldots $ ..is\\

(A) –20
(B) 20
(C) –30
(D) 30

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

If the common difference of an AP is 5, then what is a\textsubscript{18} - a\textsubscript{13}?
(A) 5
(B) 20
(C) 25
(D) 30

Answer:

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

What is the common difference of an AP in which a18 - a14 = 32?
(A) 8
(B) – 8
(C) – 4
(D) 4

Answer:

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:12

If 7 times the 7thterm of an AP is equal to 11 times its 11th term, then its 18thterm will be
(A) 7
(B) 11
(C) 18
(D) 0

Answer:

\\ \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

Answer:

\\ \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

Answer:

[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

If the first term of an AP is –5 and the common difference is 2, then the sum of the first 6 terms is
(A) 0
(B) 5
(C) 6
(D) 15

Answer:

\\$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

Answer:

\\$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

\text{In an AP if a = 1,} a\textsubscript{n} = 20 $ and $ S\textsubscript{n} = 399\text{, then n is }
(A) 19
(B) 21
(C) 38
(D) 42

Answer:

\\$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

Answer:

\\$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.

NCERT Exemplar Class 10 Maths Solutions Chapter 5 Arithmetic Progression: Exercise- 5.2

Question:1

i)Which of the following form an AP? Justify your answer.
–1, –1, –1, –1, ...
ii)
Which of the following form an AP? Justify your answer.
0, 2, 0, 2, ...
iii)
Which of the following form an AP? Justify your answer.
1, 1, 2, 2, 3, 3,...
iv)
Which of the following form an AP? Justify your answer.
11, 22, 33,...
v)
Which of the following form an AP? Justify your answer.
\frac{1}{2}, \frac{1}{3}, \frac{1}{4},......................
vi)
Which of the following form an AP? Justify your answer.
2, 2^2, 2^3, 2^4, ...
vii)
Which of the following form an AP? Justify your answer.
\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots

Answer:

i)
Answer. [True]
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)
Answer. [No]
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)
Answer. [No]
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)
Answer. [Yes]
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)
Answer. [No]
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)
Answer. [No]
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)
Answer. [Yes]
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.

Answer:

Answer. [True]

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: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?

Answer:

Answer. [True]
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.

Answer:

Answer. [False]
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.

Answer:

Answer. [True]

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.

Answer:

i)
Answer. [Yes]
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)
Answer. [Yes]
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)
Answer. [Yes]
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)
Answer. [No]
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

Answer:

i)

Answer. [Yes]

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)
Answer. [No]
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)
Answer. [No]
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.}

NCERT Exemplar Class 10 Maths Solutions Chapter 5 Arithmetic Progression: Exercise- 5.3

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


Answer:

\\(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:4

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

Answer:

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:9

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

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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:16

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

Answer:

The given AP : is - 2, - 4, -6, $ \ldots $ $ \ldots $ . , - 100\\
AP from end is - 100, - 98$ \ldots $ $ \ldots $ $ \ldots $ .. 4, -6, -2\\
\\ 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:19

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

Answer:

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: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

Answer:

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: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}

Answer:

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:32

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

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: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.

Answer:

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: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.

Answer:

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

Answer:

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: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?

Answer:1

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?

Answer:

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 Exemplar Solutions for Other Subjects:

NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

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.

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Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

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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Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

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.

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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)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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

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