# NCERT Exemplar Solution for CBSE Class 10 Mathematics: Arithmetic Progressions (Part-II)

In this article you will get CBSE Class 10 Mathematics chapter 5, Arithmetic Progressions: NCERT Exemplar Problems and Solutions (Part-II). Every question has been provided with a detailed solution. All the questions given in this article are very important to prepare for CBSE Class 10 Board Exam 2017-2018.

Here you get the CBSE Class 10 Mathematics chapter 5, Arithmetic Progressions: NCERT Exemplar Problems and Solutions (Part-II). This part of the chapter includes solutions for Exercise 5.2 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Arithmetic Progressions. This exercise comprises of only the Very Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

**Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Arithmetic Progressions:**

**Exercise – 5.2**

**Very Short Answer Type Questions**

**Question. 1** Which of the following form of an AP? Justify your answer.

**Solution.**

A series of numbers will form an AP, if d_{1} = d_{2 }= d_{3}…

(i) −1, −1, −1, −1, ….

Here, t_{1} = −1, t_{2} = −1, t_{3} = −1 and t_{4} = − 1

Now, d_{1} = t_{2} – t_{1} = − 1 + 1 = 0

d_{2} = t_{3} − t_{2} = − 1 + 1 = 0

d_{3} = t_{4} − t_{3} = − 1 + 1 = 0

As d_{1} = d_{2 }= d_{3}, therefore given list of numbers forms an AP.

(ii) 0, 2, 0, 2, …

Here, t_{1} = 0, t_{2} = 2, t_{3} = 0 and t_{4} = 2

Now, d_{1} = t_{2} − t_{1 }= 2 − 0 = 2

d_{2} = t_{3} − t_{2} = 0 − 2 = − 2

d_{3} = t_{4} − t_{3} = 2 − 0 = 2

As d_{1} ≠ d_{2}, therefore given list of numbers does not form an AP.

(iii) 1, 1, 2, 2, 3, 3, …..

Here, t_{1} = 1, t_{2} = 1, t_{3} = 2 and t_{4} = 2

Now, d_{1} = t_{2 }− t_{1 }= 1 − 1 = 0

d_{2} = t_{3} − t_{2} = 2 – 1 = 1

d_{3} = t_{4} − t_{3} = 2 − 2 = 0

As d_{1} ≠ d_{2}, therefore given list of numbers does not form an AP.

(iv) 11, 22, 33, …..

Here, t_{1} = 11, t_{2} = 22 and t_{3} = 33

Now, d_{1} = t_{2} − t_{1} = 22 − 11 = 11

d_{2} = t_{3} − t_{2} = 33 − 22 = 11

d_{3} = t_{4 }− t_{3 }= 33 − 22 = 11

As d_{1} = d_{2 }= d_{3}, therefore given list of numbers forms an AP.

As d_{1} ≠ d_{2}, therefore given list of numbers does not form an AP.

(vi) 2, 2^{2}, 2^{3}, 2^{4},…. i.e., 2, 4, 8, 16, ……

Here, t_{1} = 2, t_{2} = 4, t_{3} = 8 and t_{4} = 16

Now, d_{1} = t_{2} − t_{1} = 4 − 2 = 2

And d_{2} = t_{3} − t_{2} = 8 − 4 = 4

As d_{1} ≠ d_{2}, therefore given list of numbers does not form an AP.

As d_{1} = d_{2 }= d_{3}, therefore given list of numbers forms an AP.

Since, here a_{2} – a_{1} = a_{3} – a_{2} but a_{3} – a_{2} ≠ a_{4} – a_{3}, i.e., d_{2} ≠ d_{3 }therefore given series of numbers does not form an AP.

**Question. 3** For the AP −3, −7, −11,….. can we find directly a_{30} – a_{20} without actually finding a_{30} and a_{20}? Give reason for your answer.

**Solution.**

Yes.

We know that nth term of an AP ia given as,

a_{n} = a + (n − 1) d

Therefore, a_{30} = a + (30 − 1) d = a + 29d

And a_{20} = a + (20 - 1) d = a + 19d

⟹ a_{30} – a_{20} = (a + 29d) – (a + 19d) = 10d

Now, form given AP common difference, d = −7 – (−3) = −7 + 3 = −4

Therefore, a_{30} – a_{20} = 10 (–4) = –40

**Question. 4** Two AP’s have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10^{th} terms is the same as the difference between their 21^{st} terms, which is the same as the difference any two corresponding terms. Why?

**Solution.**

Let the same common difference of two APs be d.

Given, first term of first AP, a_{1} = 2

And, first term of second AP, a_{1}’= 7

Using, a_{n} = a + (n − 1)d,

10^{th} term of first AP, a_{10 }= 2 + 9d

And 10^{th} term of second AP, a_{10}’ =7 + 9d

So, a_{10 }− a_{10}’ = (2 + 9d) − (7 + 9d) = −5

Again using a_{n} = a + (n − 1)d,

21^{st} term of first AP, a_{21} = 2 + 20d

And 21^{st} term of second AP, a_{21}’ = 7 + 20d

So, a_{21 }− a_{21}’ = (2 + 9d) − (7 + 20d) = −5

Also, if the a_{n} and a_{n}’ are the nth terms of first and second AP.

Then, a_{n} – a_{n}’ = [2 + (n – 1)d] – [7 + (n – 1)d] = –5

Hence, the difference between any two corresponding terms of such AP’s is the same as the difference between their10^{th} and 21^{st} terms.

**Question. 5** Is 0 a term of the AP 31, 28, 25,…? Justify your answer.

**Solution.**

Let 0 be the nth term of given AP, i.e., a_{n} = 0.

Now, for given AP,

First term, a = 31

And common difference, d = 28 – 31 = −3

As nth terms of an AP is given as,

a_{n} = a + (n − 1)d

According to question,

Since, n comes out to be fraction but not a positive integer. Therefore, 0 cannot be a term of the given AP.

**Question. 6** The taxi fare 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? Given Reasons.

**Solution.**

No, the statement is not ture as the total fare (in Rs.) after each km is given as:

15, (15 + 8), (15 + 2 × 8), (15 + 3 × 8),… . = 15, 23, 31, 39,…

Let t_{1} = 15, t_{2} = 23, t_{3} = 31 and t_{4} = 39

Now, d_{1} = t_{2} – t_{1} = 23 – 5 = 8

d_{2 }= t_{3} – t_{2} = 33 – 23 = 8

And d_{3 }= t_{4} – t_{3} = 39 – 31 = 8

As d_{1} = d_{2 }= d_{3}, therefore given list of numbers formed by taking total fares after each Km forms an AP.

**Question. 7** In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.

(i) The fee charged from a student every month by a school for the whole session, when the monthly is Rs. 400.

(ii) 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 increase by Rs. 50 for the next higher class.

(iii) 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) The number of bacteria in certain food item after each second, when they double in every second.

**Solution.**

The fee charged from a student every month by a school for the whole session is

400, 400, 400, 400,…

As the difference between successive terms of given list of numbers is same as 0, so it forms an AP, with common difference, d = 0.

(ii) The monthly fee from I to XII is given as:

250, (250+50), (250 + 2 × 50), (250 + 3 × 50),….

i.e., 250, 300, 350, 400,….

Here, d_{1} = 300 – 250 = 50

d_{2} = 350 – 300 = 50

d_{3} = 400 – 350 = 50

As, d_{1} = d_{2 }= d_{3}, therefore given list of numbers formed by monthly fees from I to XII forms an AP.

So, the amount of money in the account of Varun at end of every year is given as:

1000, (1000 + 100 × 1), (1000 + 100 × 2), (1000 + 100 × 3),…..

i.e., 1000, 1100, 1200, 1300,….

Here, d_{1} = 1100 – 1000 = 100

d_{2} = 1200 – 1100 = 100

d_{3} = 1300 – 1200 = 100

As, d_{1} = d_{2 }= d_{3}, therefore given list of numbers forms an AP.

(iv) Let the number of bacteria present in food initially = *x*

Since, they double in every second.

Therefore, number of bacteria after eavery second are given as:

*x*, 2*x*, 2 (2*x*), 2 (2·2·*x*),…..

i.e., *x*, 2*x*, 4*x*, 8*x*,……

Here, d_{1} = 2*x* – *x* = *x*

d_{2} = 4*x *– 2*x* = 2*x*

d_{2} = 8*x* – 4*x* = 4*x*

As, d_{1} = d_{2 }= d_{3}, therefore given list of numbers forms an AP.

**Question. 8** Justify whether it is true to say that the following are the nth terms of an AP.

(i) 2n – 3

(ii) 3n^{2} + 5

(iii) 1 + n + n^{2}

**Solution.**

**Concept used:**To prove for the given term to be nth term of an AP, the series formed by that term must have same common difference.

(i) Yes.

Here, a_{n} = 2n – 3

⟹ a_{1} = 2(1) – 3 = −1

a_{2} = 2(2) – 3 = 1

a_{3} = 2(3) – 3 = 3

a_{4} = 2(4) – 3 = 5

Now, d_{1} = a_{2} – a_{1} = 1 – (–1) = 1 + 1 = 2

d_{2} = a_{3} – a_{2} = 3 – 1 = 2

d_{3} = a_{4} – a_{3} = 5 – 3 = 2

As, d_{1} = d_{2 }= d_{3}, therefore list of numbers −1, 1, 3, 5,…. forms an AP.

Thus, 2n – 3 is the nth term of an AP.

(ii) No.

Here a_{n} = 3n^{2} + 5

⟹ a_{1} = 3(1)^{2} + 5 = 8

⟹ a_{2} = 3(2)^{2} + 5 = 3(4) + 5 = 17

⟹ a_{3} = 3(3)^{2} + 5 = 3(9) + 5 = 27 + 5 = 32

⟹ a_{4} = 3(4)^{2} + 5 = 3(16) + 5 = 48 + 5 = 53

Now, d_{1} = a_{2} – a_{1} = 17 – 8 = 9

d_{2} = a_{3} – a_{2} = 32 – 17 = 15

d_{3} = a_{4} – a_{3} = 53 – 32 = 21

As, d_{1} ≠ d_{2 }≠ d_{3}, therefore list of numbers 8, 17, 32, 53,……does not form an AP.

Thus, 3n^{2} + 5 is not the nth term of an AP.

(iii) No.

Here a_{n} = 1 + n + n^{2}

⟹ a_{1} = 1 + 1 + (1)^{2} = 3

⟹ a_{2} = 1 + 2 + (2)^{2} = 1 + 2 + 4 = 7

⟹ a_{3} = 1 + 3 + (3)^{2} = 1 + 3 + 9 = 13

⟹ a_{4} = 1 + 3 + (4)^{2} = 1 + 4 + 16 = 21

Now, d_{1} = a_{2} – a_{1} = 7 – 3 = 4

d_{2 }= a_{3} – a_{2} = 13 – 7 = 6

d_{3} = a_{4} – a_{3} = 21 – 13 = 8

As, d_{1} ≠ d_{2 }≠ d_{3}, therefore list of numbers 3, 7, 13, 21,…. does not form an AP.

Thus, 1 + n + n^{2} is not the nth term of an AP.

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