**NCERT Solutions of class X important questions of chapter 5 ‘Arithmetic progression’**

The** arithmetic progression chapter 5** **class X** is one of the **important** part of the algebra in **NCERT class X** **mathematics**. The **arithmetic progression** is used to predict a particular quantity which is varying in a pattern, in space, universe, business, and science. The standard form of an** arithmetic progression** is a, a +d, a+ 2d, a+3d ……….where a is the first term of **A.P** and **d** is a **common difference**. Here you can study 20 most **important questions** which we have selected from the** chapter number 5** keeping in view your preparation of **board exams 2020.**

## NCERT Solutions of class X important questions of chapter 5 ‘Arithmetic progression’

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Q1. Find the common difference of the following A.P.

(i) 3, 3 + √2, 3 + 2√2, 3 + 3√2,……… (ii) 0,–4, –8, –12………..(iii) √2, √8, √18, √32…….

Answer. (i) The given series is 3, 3 + √2, 3 + 2√2, 3 + 3√2,………

Common difference is = 3 + √2 – 3 = √2

(ii) The given series is 0,–4, –8, –12………..

Common difference is = –4 –0 = –4

(iii) The given series is √2, √8, √18, √32…….

Common difference = √8 – √2 =2√2 – √2 = √2

Q2. In the following AP ‘s find the missing term in the boxes.

Answer.

a = 2, , Applying the AP formula of n th term

26 = 2 + 2d

d = 12

Missing term is second

Answer.

We are given the first term of the AP,a= 5 and 4 th term of AP =

Missing terms are second and third

Therefore the second missing term is and third missing term is 8.

Second and sixth terms of the series are 38 and –22 respectively

Applying the n th term formula of AP

a +d = 38……….(i)

a + 5d = –22……(ii)

Substracting from (i) to (ii)

–4d = 60

d = –15

Putting the value of d =–15 in eq.(i)

a –15 = 38

a = 53

Therefore first missing term is = 53

Third missing place = a +2d = 53 –2×15= 23

Fourth missing term = 23 –15 = 8

Fifth missing term = 8–15 = –7

Q3. Find the number of terms in the following AP

Let–47 is n th term of AP

Applying n th term formula of AP

a = 18,

n = 27

Hence the number of terms in the given AP is 27.

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Q4. Check whether –150 is a term of the AP 11,8,5,2 ………

a = 11, d = 8 –11 = –3,

Applying the nth term formula

–150 = 11 + (n –1) ×–3

–161 = –3(n –1)

n –1 =

Since n is not a integer so –150 can’t be a term of the given AP.

Q5. An AP consist of 50 terms of which 3 rd term is 12 and the last term is 106. Find the 29 th term.

Answer. In the given AP n = 50, third term, = 12, last term, =106

Applying the n term formula

a + 2d = 12 ……(i)

a + 49d = 106…..(ii)

Subtracting eq.(i) from (ii),we have

47d = 94

d = 2, putting this value in equation (i)

a + 4 = 12 ⇒a = 8

Therefore 29 th term of the given AP is 64

Q7. If 17 th term of an AP exceeds its 10 th term by 7 . Find the common difference.

Answer. As we know the n th term of an AP is = a + (n –1)d

According to question

d = 1

Hence the common difference of the given AP is 1.

Q8. Which term of the AP 3,15,27,39……..will be 132 more than its 54 th term.

Answer. n th term of an AP is given by the following formula

In the given AP , a = 3, d = 15 –3 = 12

Let m th term , =639 of the given AP which is 132 more than its 54 th term

3 + 12(m– 1) = 639 +132 = 771

12(m– 1) = 771 – 3 = 768

m = 65

Hence the 65 th term of the given AP is 132 more than its 54 th term.

Q9. Two AP’s have the same common difference . The difference between their 100 term is 100 . What is the difference between their 1000 th term.

Answer. Let the common difference of both AP’s = d and their first term are a and a’ So the AP’s will be a , a +d, a +2d,………, and a’, a’ +d, a’ +2d,……..

We are given that

Applying the n th term formula of an AP

Subtracting eq.(ii) from eq. (i)

a – a’ = 100……….,(iii)

Thousandth terms of both AP’s will be as follows

The difference of their thousand terms are

From eq.(iii) we have

Therefore the difference of their 1000 th terms is 100.

Q10.How many three-digit numbers are divisible by 7.

Answer. The least three-digit number is 100 ,if we divide it by 7 the remainder is 2 subtracting the remainder from the divisor 7 we get 7 –2= 5, adding 5 to 100 it comes out 105 the smallest three-digit number divisible by 7.

Largest 3 digits number is 999, dividing it by 7 we get 5 as a remainder, so the largest 3 digit number divisible by 7 will be 999 –5 = 994.

Therefore the Required AP is 105, 112, 119, 126,…….994.

Let 994 is the n th term of this AP

The n th term of the AP is given by the following formula

a = 105, d = 7, n th term is 994

994 = 105 + (n –1)7

7(n –1) = 994 – 105

7(n –1) = 889

n = 128

Hence the number of three digits numbers divisible by 7 is 128.

Q11. For what value of n , are the n th terms of two APs 63,65,67,…..and 3,10,17,….. are equal.

Answer. n th terms of AP is given as

In the first AP a = 63, d = 65 – 63 = 2,

n th term of first AP = 63 + (n – 1)2

In the second AP 3,10,17,……, a = 3, d = 10–3= 7

n th term of second AP = 3 + (n – 1)7

According question the n th term of both APs are equal

63 + (n – 1)2 = 3 + (n – 1)7

(n – 1)2 – (n – 1)7 = 3 – 63

(n – 1)(2 –7) = –60

n – 1 = 12

n = 13

Hence 13 th term of both the given AP are equal.

Q12. Find the 20 th term from the last term of the AP 3, 8, 13,…….,253.

Answer. The given AP is 3, 8, 13,…….,253.

Since we have to determine 20 th term from the last term ,so reversing the series as following.

253,248……….13,8,3

n th term of the AP is given by the following formula

a = 253, d = 248 –253 =–5, n = 20

Hence 20 th term from the last term of the given AP is 158.

Q13. The sum of 4 th and 8 th term of an AP is 24 and the sum of 6 th and 10 terms is 44. Find the first three terms of the AP.

Answer. The nth term of an AP is given by

4 th term of the given AP = a + (4 – 1)d = a +3d and 8 th term =a + (8 – 1)d=a + 7d

According to first condition

4 th term of the given AP + 8 th term= 24

a +3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12………(i)

6 th term of the given AP = a + (6– 1)d =a + 5d and 10 th term = a +(10 –1)d=a +9d

according to second condition

6 th term of the given AP + 10 th term = 44

a + 5d + a +9d = 44

2a + 14d = 44

a + 7d = 22…….(ii)

Subtracting eq.(i) from eq.(ii)

d =5

Substituting value of d in equation (i)

a + 5 ×5 =12

a = –13

d = 5 and a =–13

The value of first term is –13

Second term = –13 +5 = –8

Third term –8 + 5 = –3

Therefore the first, second and third term are –13, –8 and –3.

Q14.Find the sum of the AP to 11 terms.

Answer. The sum of 11 terms is given by

Let sum of 11 terms is

Hence the sum of the given AP up to 11 terms is .

Q15. In an AP first term is 2 , the common difference is 8 and sum of n terms is 90, find n and nth term of the AP.

Answer. First term of the AP ,a = 2, common difference,d =8 and sum of n term = 90.

We are given the sum of n terms = 90, which is given by the following formula.

180 = 8n² –4n

8n² –4n –180 = 0

2n² – n – 45 = 0

45 × 2 = 5×3×3×2 = 10×9

2n² –10n +9n – 45 = 0

2n(n – 5) + 9(n –5) = 0

(n –5)(2n +9) =0

The value of n must be positive integer ,so neglecting the second root ,therefore the value of n=5

n th term of AP = a +(n –1)d

5 th term of AP = 2 + (5 –1)8= 34

Q16. Find the sum of first 22 terms of an AP in which d =7 and 22 nd term is 149.

Answer. We are given d = 2, n = 22 and 22 nd term = 149 ,

, for getting the sum of 22 terms we are needed the value of first term ,as we are given here 22 nd term=149, therefore applying the n th term formula

149 = a + (22 – 1)7

149 = a + 147

a = 149–147 = 2

The sum of the n terms is given by the following formula.

= 11 ×151 =1661

Therefore the sum of 22 term of the given AP is 1661.

Q17. If n th term of an AP is defined as 3 +4n, then find the sum of its first 15 terms.

Answer. The n th term of the given AP is 3 + 4n

………………………..

…………………………

a = = 7, d = 11 – 7 = 4

= 15 × 35 = 525

Hence the sum of the first 15 terms is 525.

Q18. If the sum of the first n terms of an AP is 4n –n², What is the first term? What is the sum of the first two terms? What is the second term? Similarly find the third term, the tenth and nth term.

Answer. The sum of first n terms of an AP is = 4n–n²

,

………………..

……………….

Therefore the AP is 3,1,–1…….

n th term of this AP is given by the following formula

a = 3, d = 1 –3 =–2

Therefore the first term of the required AP is 3

The sum of the first two terms = 4

The second term of AP = 1

The third term of the AP= –1

Tenth term of the AP = –15

nth term of the AP = 5 –2n

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Q19. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Answer. Let the first prize is given to a student = x

Second prize = x –20

Third prize = x –40

…………

………..

It is an AP of 7 terms x, x –20, x –40….upto seven terms

a =x, d =–20 and sum of seven terms is total cash,

Sum of n terms of the AP is given by the following formula

2x –120 = 200

2x = 320

x = 160

Therefore first prize is = Rs 160, second prize =Rs 140, third prize=Rs 120, fourth prize =Rs 100, fifth prize = Rs 80, sixth prize=Rs 60, seventh prize =Rs 40.

Q20.In a school students thought of planting trees in and around the school to reduce air pollution. it was decided the no of trees that each section of each class will plant will be the same as the class in which they are studying .e.g a section of class 1 will plant 1 tree,a section of class 2 will plant 2 trees and so on till class 12.there are 3 sections of each class. how many trees will be planted by the students?

Answer. Since each class has three section ,so according to question class 1 plants, the number of trees =1 ×3 =3

Class second plant trees = 2 × 3 = 6

Class third plants trees= 3× 3 = 9

………………

……………..

Class 12 plants trees = 12× 3 = 36

Since it is an AP 3, 6,9,12……36

So, the total number of trees planted by the overall 12 classes is calculated by the following formula.

= 6(6 + 11×3) = 6×39 = 234

Hence the total number of trees planted is 234.

Q21. Which term of the AP, 121,117,113……… is its first negative term?

Answer.a = 121, d = 117 –121 = –4

The condition of the first negative term of the given AP is such that

= 121 + (n –1)(–4) = 121 –4n +4= 125 –4n

125 –4n < 0

125 < 4n

First integer greater than is 32, hence first negative term of the series is 32 th term.

Q22. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of x such that sum of the number of houses preceding the house numbered x is equal to the sum of the number of houses following x.

Answer. The houses are numbered as follows,together we are also given a house number x within 1 to 49.

1,2,3,4…….x–1,x,x +1,x+2………..47,48,49

We are given the sum of house numbers preceding the house numbered x = Sum of the house numbers followed by x

The houses from 1 to x – 1 will be as given bellow

Let S is the sum of the numbers of houses preceding the house no.x and S’ is the sum of the numbers of houses followed by x

a =1, d =1, l =x –1

And

According to question

(x –1)x = (49 – x)(x + 50)

x² –x = 49x +2450 –x² –50x

2x² = 245o

x ² =1225

x = √(1225)

x = 35

So,the house no x = 35 exist as described in the question.

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