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Wednesday, September 13, 2023

160-NCERT-10-5-Arithmetic Progressions - Ex-5.1

NCERT
10th Mathematics
Exercise 5.1
Topic: 5 Arithmetic Progressions

Click here for ⇨ NCERT-10-4-Quadratic Equations-Ex- 4.4

EXERCISE 5.1

Q1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
 
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
 
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
 
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8 % per annum.

Explanation:

1) An arithmetic progression is a list of numbers in which each term is obtained by
adding a fixed number to the preceding term except the first term.
2) The fixed number is known as the common difference of an AP.
3) The common difference may be positive, negative, or zero.
4) If the first of an AP is "a" and the common difference is "d" then the terms of an
AP will be:  a, (a + d), (a + 2d), (a + 3d) ...

Solution:

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

1) According to the problem,
a) Taxi fare for the first km = 15.
b) Taxi fare for the first 2 km = 15 + 8 = 23.
c) Taxi fare for the first 3 km = 15 + 8 + 8 = 31.
2) As the terms increment by a constant 8, it forms an AP.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

1) Let the amount of air in the cylinder be x.
2) According to the problem, every time the vacuum pump removes (1/4) of air
remaining in the cylinder, so
a) The volume after first removal
= x - (x/4)
= x(1 - 1/4)
= 3x/4 ---------------- equation 1
b) The volume after second removal 
= 3x/4 - 1/4(3x/4)
= (3x/4)(1 - 1/4)
= (3x/4)(4 - 1)/4
= (3x/4)(3/4)
= 9x/16 ---------------- equation 2
c) The volume after third removal 
= 9x/16 - 1/4(9x/16)
= (9x/16)(1 - 1/4)
= (9x/16)(3/4)
= 27x/64 ---------------- equation 3
3) Now we will check the terms x, 3x/4, 9x/16, 27x/64 are in AP or not.
a) Second term - first term = 3x/4 - x 
   = (3x - 4x)/4
   = - x/4 ---------------- equation 4.
b) Third term - Second term = 9x/16 - 3x/4 
     = (9x - 12x)/16
     = - 3x/16 ---------------- equation 5.
4) From equation 4 and equation 5, as the differences between the terms are not
same, we can say that these terms are not in AP.

(iii) The cost of digging a well after every meter of digging, when it costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter. 

1) According to the problem,
a) Cost of digging the well after 1 meter = 150.
b) Cost of digging the well after 2 meter = 150 + 50 = 200.
c) Cost of digging the well after 3 meter = 150 + 50 + 50 = 250.
2) As the terms increment by a constant 50, it forms an AP.

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8 % per annum.

1) We know that if Rs P is invested at the rate of r % per annum for n years, the
amount received will be A = P[1+(r/100)]n.
2) According to the problem, for P = 10000, r = 8 %, 
a) The amount after first year
A = P[1+(r/100)]n
A = 10000[1+(8/100)]1
A = 10000[1+(8/100)] ---------------- equation 1
b) The amount after second year 
A = P[1+(r/100)]n
A = 10000[1+(8/100)]2 ---------------- equation 2
c) The amount after third year 
A = P[1+(r/100)]n
A = 10000[1+(8/100)]3 ---------------- equation 3
3) Now we will check the terms [1+(8/100)], [1+(8/100)]2, [1+(8/100)]3, 
[1+(8/100)]4 are in AP or not.
a) Second term - first term
= 10000[1+(8/100)]2 - 10000[1+(8/100)]
= 10000[1+(8/100)]{1+(8/100) - 1}
= 10000[1+(8/100)](8/100) 
= 10000(8/100)[1+(8/100)] ---------------- equation 4.
b) Third term - Second term 
= 10000[1+(8/100)]3 - 10000[1+(8/100)]2
= 10000[1+(8/100)]2{1+(8/100) - 1}
= 10000[1+(8/100)]2(8/100) 
= 10000(8/100)[1+(8/100)]2 ---------------- equation 5.
4) From equation 4 and equation 5, as the differences between the terms are not
same, we can say that these terms are not in AP.

Q2. Write the first four terms of the AP, when the first term a, and the common difference d are given as follows:

(i) a = 10, d = 10     (ii) a = –2, d = 0     (iii) a = 4, d = – 3
 
(iv) a = – 1, d = 1/2     (v) a = – 1.25, d = – 0.25

Explanation:

1) An arithmetic progression is a list of numbers in which each term is obtained by
adding a fixed number to the preceding term except the first term.
2) The fixed number is known as the common difference of an AP.
3) The common difference of may positive, negative, or zero.
4) If the first of an AP is "a" and the common difference is "d" then the terms of an
AP will be:  a, (a + d), (a + 2d), (a + 3d) ...

Solution:

(i) a = 10, d = 10

1) Let the terms of an AP be a1, a2, a3, a4.
2) Here, a1 = a = 10 is the first term and d = 10 is the common difference.
3) Now we will find the terms of an AP:
a) First term:
a1 = 10
b) Second term:
a2 = a1 + d
a2 = 10 + 10
a2 = 20
c) Third term:
a3 = a2 + d
a3 = 20 + 10
a3 = 30
d) Fourth term:
a4 = a3 + d
a4 = 30 + 10
a4 = 40
4) So the first 4 terms of an AP with a =10 and d = 10 are 10, 20, 30, and 40.

(ii) a = –2, d = 0

1) Let the terms of an AP be a1, a2, a3, a4.
2) Here, a1 = a = - 2 is the first term and d = 0 is the common difference.
3) Now we will find the terms of an AP:
a) First term:
a1 = - 2
b) Second term:
a2 = a1 + d
a2 = - 2 + 0
a2 = - 2
c) Third term:
a3 = a2 + d
a3 = - 2 + 0
a3 = - 2
d) Fourth term:
a4 = a3 + d
a4 = - 2 + 0
a4 = - 2
4) So first 4 terms of an AP with a = - 2 and d = 0 are - 2, - 2, - 2, and - 2.

(iii) a = 4, d = – 3

1) Let the terms of an AP be a1, a2, a3, a4.
2) Here, a1 = a = 4 is the first term and d = - 3 is the common difference.
3) Now we will find the terms of an AP:
a) First term:
a1 = 4
b) Second term:
a2 = a1 + d
a2 = 4 - 3
a2 = 1
c) Third term:
a3 = a2 + d
a3 = 1 - 3
a3 = - 2
d) Fourth term:
a4 = a3 + d
a4 = - 2 - 3
a4 = - 5
4) So first 4 terms of an AP with a = 4 and d = - 3 are 4, 1, - 2, and - 5.

(iv) a = – 1, d = 1/2

1) Let the terms of an AP be a1, a2, a3, a4.
2) Here, a1 = a = - 1 is the first term and d = 1/2 is the common difference.
3) Now we will find the terms of an AP:
a) First term:
a1 = - 1
b) Second term:
a2 = a1 + d
a2 = - 1 + 1/2
a2 = - 1/2
c) Third term:
a3 = a2 + d
a3 = - 1/2 + 1/2
a3 = 0
d) Fourth term:
a4 = a3 + d
a4 = 0 + 1/2
a4 = 1/2
4) So first 4 terms of an AP with a = - 1 and d = 1/2 are - 1, - 1/2, 0, and 1/2.

(v) a = – 1.25, d = – 0.25

1) Let the terms of an AP be a1, a2, a3, a4.
2) Here, a1 = a = - 1.25 is the first term and d = - 0.25 is the common difference.
3) Now we will find the terms of an AP:
a) First term:
a1 = - 1.25
b) Second term:
a2 = a1 + d
a2 = - 1.25 - 0.25
a2 = - 1.50
c) Third term:
a3 = a2 + d
a3 = - 1.50 - 0.25
a3 = - 1.75
d) Fourth term:
a4 = a3 + d
a4 = - 1.75 - 0.25
a4 = - 2.00
4) So first 4 terms of an AP with a = - 1.25 and d = - 0.25 are - 1.25, - 1.50, - 1.75,
and - 2.00.

Q3. For the following APs, write the first term and the common difference:
(i) 3, 1, – 1, – 3, . . .     (ii) – 5, – 1, 3, 7, . . .
 
(iii) 1/3, 5/3, 9/3, 13/3, . . .     (iv) 0.6, 1.7, 2.8, 3.9, . . .

Explanation:

1) An arithmetic progression is a list of numbers in which each term is obtained by
adding a fixed number to the preceding term except the first term.
2) The fixed number is known as the common difference of an AP.
3) The common difference may be positive, negative, or zero.
4) If the first of an AP is "a" and the common difference is "d" then the terms of an
AP will be:  a, (a + d), (a + 2d), (a + 3d) ...
5) For the terms a1, a2, a3, a4, when a2 - a1 = a3 - a2 = a4 - a3 = d, then we say
that the terms a1, a2, a3, a4, are in AP.

Solution:

(i) 3, 1, – 1, – 3, . . .

1) Here the first term is a = 3.
2) Here a1 = a = 3, a2 = 1, a3 = - 1, a4 = - 3.
3) Here, 
a) First difference:
d = a2 - a1 = 1 - 3
d = a2 - a1 = - 2 --------- equation 1
b) Second difference:
d = a3 - a2 = - 1 - 1
d = a3 - a2 = - 2 --------- equation 2
4) Here the first term a = 3 and the common difference d = - 2.  

(ii) – 5, – 1, 3, 7, . . .

1) Here the first term is a = - 5.
2) Here a1 = a = - 5, a2 = - 1, a3 = 3, a4 = 7.
3) Here, 
a) First difference:
d = a2 - a1 = - 1 - (- 5)
d = a2 - a1 = - 1 + 5 
d = a2 - a1 = 4 --------- equation 1
b) Second difference:
d = a3 - a2 = 3 - (- 1)
d = a3 - a2 = 3 + 1 
d = a3 - a2 = 4 --------- equation 2
4) Here the first term a = - 5 and the common difference d = 4.

(iii) 1/3, 5/3, 9/3, 13/3, . . .

1) Here the first term is a = 1/3.
2) Here a1 = a = 1/3, a2 = 5/3, a3 = 9/3, a4 = 13/3.
3) Here, 
a) First difference:
d = a2 - a1 = 5/3 - 1/3
d = a2 - a1 = (5 - 1)/3 
d = a2 - a1 = 4/3 --------- equation 1
b) Second difference:
d = a3 - a2 = 9/3 - 5/3
d = a3 - a2 = (9 - 5)/3 
d = a3 - a2 = 4/3 --------- equation 2
4) Here the first term a = 1/3 and the common difference d = 4/3.

(iv) 0.6, 1.7, 2.8, 3.9, . . .

1) Here the first term is a = 0.6.
2) Here a1 = a = 0.6, a2 = 1.7, a3 = 2.8, a4 = 3.9.
3) Here, 
a) First difference:
d = a2 - a1 = 1.7 - 0.6
d = a2 - a1 = 1.1 --------- equation 1
b) Second difference:
d = a3 - a2 = 2.8 - 1.7
d = a3 - a2 = 1.1 --------- equation 2
4) Here the first term a = 0.6 and the common difference d = 1.1.

Q4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, . . .     
(ii) 2, 5/2, 3, 7/2, . . .      
(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . 
(iv) – 10, – 6, – 2, 2, . . .
(v) 3, 3 + √2 , 3 + 2 √2 , 3 + 3 √2 , . . .
(vi) 0.2, 0.22, 0.222, 0.2222, . . .
(vii) 0, – 4, – 8, –12, . . .
(viii) - 1/2, - 1/2, - 1/2, - 1/2, . . .
(ix) 1, 3, 9, 27, . . .
(x) a, 2a, 3a, 4a, . . .
(xi) a, a2, a3, a4, . . .
(xii) √2, √8, √18 , √32, . . .
(xiii) √3, √6, √9 , √12 , . . .
(xiv) 12, 32, 52, 72, . . .
(xv) 12, 52, 72, 73, . . .

Explanation:

1) An arithmetic progression is a list of numbers in which each term is obtained by
adding a fixed number to the preceding term except the first term.
2) The fixed number is known as the common difference of an AP.
3) The common difference may be positive, negative, or zero.
4) If the first of an AP is "a" and the common difference is "d" then the terms of an
AP will be:  a, (a + d), (a + 2d), (a + 3d) ...
5) For the terms a1, a2, a3, a4, when a2 - a1 = a3 - a2 = a4 - a3 = d, then we say
that the terms a1, a2, a3, a4, are in AP.

Solution:

(i) 2, 4, 8, 16, . . . 

1) Here the first term is a = 2.
2) Here a1 = a = 2, a2 = 4, a3 = 8, a4 = 16.
3) Here, 
a) First difference:
d = a2 - a1 = 4 - 2
d = a2 - a1 = 2 --------- equation 1
b) Second difference:
d = a3 - a2 = 8 - 4
d = a3 - a2 = 4 --------- equation 2
4) From equation 1 and equation 2, a2 - a1 ≠ a3 - a2, so given terms are not in AP.

(ii) 2, 5/2, 3, 7/2, . . .

1) Here the first term is a = 2.
2) Here a1 = a = 2, a2 = 5/2, a3 = 3, a4 = 7/2.
3) Here, 
a) First difference:
d = a2 - a1 = 5/2 - 2
d = a2 - a1 = (5 - 4)/2 
d = a2 - a1 = 1/2 --------- equation 1
b) Second difference:
d = a3 - a2 = 3 - 5/2
d = a3 - a2 = (6 - 5)/2 
d = a3 - a2 = 1/2 --------- equation 2
4) From equation 1 and equation 2, a2 - a1 = a3 - a2, so given terms are in AP.
5) So, here the common difference d = 1/2.
6) So the next 3 terms are:
a) Fifth term:
a5 =  a4 + d
a5 =  7/2 + 1/2
a5 =  (7 + 1)/2
a5 =  8/2
a5 =  4
b) Sixth term:
a6 =  a5 + d
a6 =  4 + 1/2
a6 =  (8 + 1)/2
a6 =  9/2
c) Seventh term:
a7 =  a6 + d
a7 =  9/2 + 1/2
a7 =  (9 + 1)/2
a7 =  10/2
a7 =  5
 7) The next 3 terms are 4, 9/2, and 5.
 
(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . .

1) Here the first term is a = - 1.2.
2) Here a1 = a = - 1.2, a2 = - 3.2, a3 = - 5.2, a4 = - 7.2.
3) Here, 
a) First difference:
d = a2 - a1 = (- 3.2) - (- 1.2)
d = a2 - a1 = - 3.2 + 1.2 
d = a2 - a1 = - 2 --------- equation 1
b) Second difference:
d = a3 - a2 = (- 5.2) - (- 3.2)
d = a3 - a2 = - 5.2 + 3.2 
d = a3 - a2 = - 2 --------- equation 2
4) From equation 1 and equation 2, a2 - a1 = a3 - a2, so given terms are in AP.
5) So, here the common difference is d = - 2.
6) So the next 3 terms are:
a) Fifth term:
a5 =  a4 + d
a5 =  - 7.2 + (- 2)
a5 =  - 7.2 - 2
a5 =  - 9.2
b) Sixth term:
a6 =  a5 + d
a6 =  - 9.2 + (- 2)
a6 =  - 9.2 - 2
a6 =  - 11.2
c) Seventh term:
a7 =  a6 + d
a7 =  - 11.2 + (- 2)
a7 =  - 11.2 - 2
a7 =  - 13.2
 7) So the next 3 terms are - 9.2, - 11.2, and - 13.2.

(iv) – 10, – 6, – 2, 2, . . .

1) Here the first term is a = - 10.
2) Here a1 = a = - 10, a2 = - 6, a3 = - 2, a4 = 2.
3) Here, 
a) First difference:
d = a2 - a1 = (- 6) - (- 10)
d = a2 - a1 = - 6 + 10 
d = a2 - a1 = 4 --------- equation 1
b) Second difference:
d = a3 - a2 = (- 2) - (- 6)
d = a3 - a2 = - 2 + 6 
d = a3 - a2 = 4 --------- equation 2
4) From equation 1 and equation 2, a2 - a1 = a3 - a2, so given terms are in AP.
5) So, here the common difference is d = 4.
6) So the next 3 terms are:
a) Fifth term:
a5 =  a4 + d
a5 =  2 + 4
a5 =  6
b) Sixth term:
a6 =  a5 + d
a6 =  6 + 2
a6 =  8
c) Seventh term:
a7 =  a6 + d
a7 =  8 + 2
a7 =  10
 7) The next 3 terms are 6, 8, and 10.

(v) 3, 3 + √2 , 3 + 2√2 , 3 + 3 √2 , . . .

1) Here the first term is a = 3.
2) Here a1 = a = 3, a2 = 3 + √2, a3 = 3 + 2√2, a4 = 3 + 3√2.
3) Here, 
a) First difference:
d = a2 - a1 = (3 + √2) - (3)
d = a2 - a1 = 3 + √2 - 3 
d = a2 - a1 = √2 --------- equation 1
b) Second difference:
d = a3 - a2 = (3 + 2√2) - (3 + √2)
d = a3 - a2 = 3 + 2√2 - 3 - √2
d = a3 - a2 = √2 --------- equation 2
4) From equation 1 and equation 2, a2 - a1 = a3 - a2, so given terms are in AP.
5) So, here the common difference is d = √2.
6) So the next 3 terms are:
a) Fifth term:
a5 =  a4 + d
a5 =  (3 + 3√2) + √2
a5 =  3 + 4√2
b) Sixth term:
a6 =  a5 + d
a6 =  (3 + 4√2) + √2
a6 =  3 + 5√2
c) Seventh term:
a7 =  a6 + d
a6 =  (3 + 5√2) + √2
a6 =  3 + 6√2
 7) So next 3 terms are (3 + 4√2), (3 + 5√2), and (3 + 6√2).

(vi) 0.2, 0.22, 0.222, 0.2222, . . .

1) Here the first term is a = 0.2.
2) Here a1 = a = 0.2, a2 = 0.22, a3 = 0.222, a4 = 0.2222.
3) Here, 
a) First difference:
d = a2 - a1 = (0.22) - (0.2)
d = a2 - a1 = 0.22 - 0.2 
d = a2 - a1 = 0.02 --------- equation 1
b) Second difference:
d = a3 - a2 = (0.222) - (0.22)
d = a3 - a2 = 0.222 - 0.22 
d = a3 - a2 = 0.002 --------- equation 2
4) From equation 1 and equation 2, a2 - a1 ≠ a3 - a2, so given terms are not in AP.

(vii) 0, – 4, – 8, –12, . . .

1) Here the first term is a = 0.
2) Here a1 = a = 0, a2 = - 4, a3 = - 8, a4 = - 12.
3) Here, 
a) First difference:
d = a2 - a1 = (- 4) - (0)
d = a2 - a1 = - 4 + 0 
d = a2 - a1 = - 4 --------- equation 1
b) Second difference:
d = a3 - a2 = (- 8) - (- 4)
d = a3 - a2 = - 8 + 4 
d = a3 - a2 = - 4 --------- equation 2
4) From equation 1 and equation 2, a2 - a1 = a3 - a2, so given terms are in AP.
5) So, here the common difference is d = - 4.
6) So the next 3 terms are:
a) Fifth term:
a5 =  a4 + d
a5 =  - 12 + (- 4)
a5 =  - 16
b) Sixth term:
a6 =  a5 + d
a6 =  - 16 + (- 4)
a6 =  - 20
c) Seventh term:
a7 =  a6 + d
a7 =  - 20 + (- 4)
a7 =  - 24
 7) The next 3 terms are - 16, - 20, and - 24.

(viii) - 1/2, - 1/2, - 1/2, - 1/2, . . .

1) Here the first term is a = - 1/2.
2) Here a1 = a = - 1/2 , a2 = - 1/2, a3 = - 1/2, a4 = - 1/2.
3) Here, 
a) First difference:
d = a2 - a1 = (- 1/2) - (- 1/2)
d = a2 - a1 = - 1/2 + 1/2 
d = a2 - a1 = 0 --------- equation 1
b) Second difference:
d = a3 - a2 = (- 1/2) - (- 1/2)
d = a3 - a2 = - 1/2 + 1/2  
d = a3 - a2 = 0 --------- equation 2
4) From equation 1 and equation 2, a2 - a1 = a3 - a2, so given terms are in AP.
5) So, here the common difference is d = 0.
6) So the next 3 terms are:
a) Fifth term:
a5 =  a4 + d
a5 =  - 1/2 + 0
a5 =  - 1/2
b) Sixth term:
a6 =  a5 + d
a6 =  - 1/2 + 0
a6 =  - 1/2
c) Seventh term:
a7 =  a6 + d
a7 =  - 1/2 + 0
a7 =  - 1/2
 7) So next 3 terms are - 1/2, - 1/2, and - 1/2.

(ix) 1, 3, 9, 27, . . .

1) Here the first term is a = 1.
2) Here a1 = a = 1 , a2 = 3 a3 = 9, a4 = 27.
3) Here, 
a) First difference:
d = a2 - a1 = (3) - (1)
d = a2 - a1 = 3 - 1 
d = a2 - a1 = 2 --------- equation 1
b) Second difference:
d = a3 - a2 = 9 - 3
d = a3 - a2 = 6 --------- equation 2
4) From equation 1 and equation 2, a2 - a1 ≠ a3 - a2, so given terms are not in AP.

(x) a, 2a, 3a, 4a, . . .

1) Here the first term is a = a.
2) Here a1 = a = a, a2 = 2a, a3 = 3a, a4 = 4a.
3) Here, 
a) First difference:
d = a2 - a1 = (2a) - (a)
d = a2 - a1 = 2a - a
d = a2 - a1 = a --------- equation 1
b) Second difference:
d = a3 - a2 = (3a) - (2a)
d = a3 - a2 = 3a - 2a 
d = a3 - a2 = a --------- equation 2
4) From equation 1 and equation 2, a2 - a1 = a3 - a2, so given terms are in AP.
5) So, here common difference d = a.
6) So the next 3 terms are:
a) Fifth term:
a5 =  a4 + d
a5 =  4a + a
a5 =  5a
b) Sixth term:
a6 =  a5 + d
a6 =  5a + a
a6 =  6a
c) Seventh term:
a7 =  a6 + d
a7 =  6a + a
a7 =  7a
 7) The next 3 terms are 5a, 6a, and 7a.

(xi) a, a2, a3, a4, . . .

1) Here the first term is a = a.
2) Here a1 = a = a, a2 = a2, a3 = a3, a4 = a4.
3) Here, 
a) First difference:
d = a2 - a1 = (a2) - (a)
d = a2 - a1 = a(a - 1) --------- equation 1
b) Second difference:
d = a3 - a2 = (a3) - (a2)
d = a3 - a2 = a2(a - 1) --------- equation 2
4) From equation 1 and equation 2, a2 - a1 ≠ a3 - a2, so given terms are not in AP.

(xii) √2, √8, √18 , √32, . . .

1) Here the first term is a = √2.
2) Here a1 = a = √2, a2 = √8, a3 = √18, a4 = √32.
3) Here, 
a) First difference:
d = a2 - a1 = √8 - √2
d = a2 - a1 = 2√2 - √2
d = a2 - a1 = √2 --------- equation 1
b) Second difference:
d = a3 - a2 = √18 - √8
d = a3 - a2 = 3√2 - 2√2
d = a3 - a2 = √2 --------- equation 2
4) From equation 1 and equation 2, a2 - a1 = a3 - a2, so given terms are in AP.
5) So, here the common difference is d = √2.
6) So the next 3 terms are:
a) Fifth term:
a5 =  a4 + d
a5 =  √32 + √2
a5 =  4√2 + √2
a5 =  5√2
a5 =  √50 
b) Sixth term:
a6 =  a4 + d
a6 =  √50 + √2
a6 =  5√2 + √2
a6 =  6√2
a6 =  √72 
c) Seventh term:
a7 =  a4 + d
a7 =  √72 + √2
a7 =  6√2 + √2
a7 =  7√2
a7 =  √98 
 7) The next 3 terms are √50, √72, and √98.

(xiii) √3, √6, √9 , √12 , . . .

1) Here the first term is a = √3.
2) Here a1 = a = √3, a2 = √6, a3 = √9, a4 = √12.
3) Here, 
a) First difference:
d = a2 - a1 = √6 - √3
d = a2 - a1 = √3√2 - √3
d = a2 - a1 = √3(√2 - 1) --------- equation 1
b) Second difference:
d = a3 - a2 = √9 - √6
d = a3 - a2 = √3√3 - √3√2
d = a3 - a2 = √3(√3 - √2) --------- equation 2 
4) From equation 1 and equation 2, a2 - a1 ≠ a3 - a2, so given terms are not in AP.

(xiv) 12, 32, 52, 72, . . .

1) Here the first term is a = 12.
2) Here a1 = a = 12, a2 = 32, a3 = 53, a4 = 74.
3) Here, 
a) First difference:
d = a2 - a1 = 32 - 12
d = a2 - a1 = 9 - 1 
d = a2 - a1 = 8 --------- equation 1
b) Second difference:
d = a3 - a2 = 52 - 32
d = a3 - a2 = 25 - 9 
d = a3 - a2 = 16 --------- equation 2
4) From equation 1 and equation 2, a2 - a1 ≠ a3 - a2, so given terms are not in AP.

(xv) 12, 52, 72, 73, . . .

1) Here the first term is a = 12.
2) Here a1 = a = 12, a2 = 52, a3 = 72, a4 = 73.
3) Here, 
a) First difference:
d = a2 - a1 = 52 - 12
d = a2 - a1 = 25 - 1 
d = a2 - a1 = 24 --------- equation 1
b) Second difference:
d = a3 - a2 = 72 - 52
d = a3 - a2 = 49 - 25 
d = a3 - a2 = 24 --------- equation 2
c) Third difference:
d = a4 - a3 = 73 - 72
d = a4 - a3 = 73 - 49
d = a4 - a3 = 24 --------- equation 3
4) From equation 1, 2 and 3, a2 - a1 = a3 - a2 = a4 - a3, so given terms are in AP.
5) So, here the common difference is d = 24.
6) So the next 3 terms are:
a) Fifth term:
a5 = a4 + d
a5 = 73 + 24
a5 = 97 
b) Sixth term:
a6 =  a4 + d
a6 = 97 + 24
a6 = 121 
c) Seventh term:
a7 =  a4 + d
a7 = 121 + 24
a7 = 145 
 7) The next 3 terms are 97, 121, and 145.

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Wednesday, September 6, 2023

159-NCERT-10-4-Quadratic Equations - Ex-4.4

NCERT
10th Mathematics
Exercise 4.4
Topic: 4 Quadratic Equations

Click here for ⇨ NCERT-10-4-Quadratic Equations-Ex- 4.3

EXERCISE 4.4

Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x2 – 3x + 5 = 0 (ii) 3x2 – 4√3 x + 4 = 0 (iii) 2x2 – 6x + 3 = 0 

Explanation:

1) For the quadratic equation is of the form ax2 + bx + c = 0, where a ≠ 0, 
(b2 - 4ac) is known as discriminant.
2) The quadratic equation ax2 + bx + c = 0 has
(a) two distinct real roots, if b2 – 4ac > 0,
(b) two equal real roots, if b2 – 4ac = 0,
(c) no real roots, if b2 – 4ac < 0.

Solution:

(i) 2x2 – 3x + 5 = 0

1) The given equation is 2x2 - 3x + 5 = 0 ------------------ equation 1.
2) Equate the coefficient of equation 2x2 - 3x + 5 = 0 with ax2 + bx + c = 0, we have,
a = 2, b = - 3, c = 5.
3) First we will find:
b2 - 4ac = (- 3)2 - 4(2)(5)
b2 - 4ac = 9 - 40
b2 - 4ac = - 31 ------------------ equation 2. 
4) As b2 - 4ac < 0, it has no real roots of the quadratic equation 2x2 - 3x + 5 = 0.

(ii) 3x2 – 4√3 x + 4 = 0

1) The given equation is 3x2 – 4√3 x + 4 = 0 ------------------ equation 1.
2) Equate the coefficient of equation 3x2 – 4√3 x + 4 = 0 with ax2 + bx + c = 0, 
we have,
a = 3, b = - 4√3, c = 4.
3) First we will find:
b2 - 4ac = (- 4√3)2 - 4(3)(4)
b2 - 4ac = 48 - 48
b2 - 4ac = 0 ------------------ equation 2. 
4) As, b2 - 4ac = 0, it has two equal real roots, so from equation 2 and equation 3,
we have,
x = [- b ± √(b2 - 4ac)]/2a
x = [- (- 4√3) ± √0]/2(3)
x = (4√3 ± 0)/2(3)
x = 2(√3)/3
x = 2(√3)/(√3√3)
x = 2/√3
 5) So,  x = 2/√3 or x = 2/√3.

(iii) 2x2 – 6x + 3 = 0

1) The given equation is 2x2 – 6x + 3 = 0 ------------------ equation 1.
2) Equate the coefficient of equation 2x2 – 6x + 3 = 0 with ax2 + bx + c = 0, 
we have,
a = 2, b = - 6, c = 3.
3) First we will find:
b2 - 4ac = (- 6)2 - 4(2)(3)
b2 - 4ac = 36 - 24
b2 - 4ac = 12 ------------------ equation 2. 
4) As, b2 - 4ac ≥ 0, it has two distinct real roots, so from equation 2 and equation 3,
we have,
x = [- b ± √(b2 - 4ac)]/2a
x = [- (- 6) ± √12]/2(2)
x = (6 ± √12)/6
x = (6 ± 2√3)/6
x = (3 ± √3)/3
5) So,  x = (3 + √3)/3 or x = (3 - √3)/3.

Q2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0

Explanation:

1) For the quadratic equation is of the form ax2 + bx + c = 0, where a ≠ 0, 
(b2 - 4ac) is known as discriminant.
2) The quadratic equation ax2 + bx + c = 0 has
(a) two equal real roots, if b2 – 4ac = 0,

Solution:

(i) 2x2 + kx + 3 = 0

1) The given equation is 2x2 + kx + 3 = 0 ------------------ equation 1.
2) Equate the coefficient of equation 2x2 + kx + 3 = 0 with ax2 + bx + c = 0, 
we have,
a = 2, b = k, c = 3.
3) First we will find:
b2 - 4ac = (k)2 - 4(2)(3)
b2 - 4ac = k2 - 24 ------------------ equation 2. 
4) As the quadratic equation has two equal roots, 
b2 - 4ac = 0
k2 - 24 = 0
k2 = 24 
k = ± √24 = ± 2√6
 5) So,  k = 2√6 or k = - 2√6.

(ii) kx (x – 2) + 6 = 0

1) The given equation is 
kx (x – 2) + 6 = 0
kx2 – 2kx + 6 = 0 ------------------ equation 1.
2) Equate the coefficient of equation kx2 – 2kx + 6 = 0 with ax2 + bx + c = 0, 
we have,
a = k, b = - 2k, c = 6.
3) First we will find:
b2 - 4ac = (- 2k)2 - 4(k)(6)
b2 - 4ac = 4k2 - 24k ------------------ equation 2. 
4) As the quadratic equation has two equal roots, 
b2 - 4ac = 0
4k2 - 24k = 0
4k(k - 6) = 0
k(k - 6) = 0
 5) So,  k = 0 or k = 6.

Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m2? If so, find its length and breadth.


1) Let the breadth of a rectangular mango grove be x m.
2) So, the length of a rectangular mango grove will be 2x m 
3) According to the problem, the area is 800, so
2x(x) = 800
2x2 = 800
x2 = 400
x2 - 400 = 0 ------------------ equation 1.
4) Equate the coefficient of equation x2 – 400 = 0 with ax2 + bx + c = 0, 
we have,
a = 1, b = 0, c = - 400.
5) First we will find:
b2 - 4ac = (0)2 - 4(1)(- 400)
b2 - 4ac = 1600 ------------------ equation 2.
6) As b2 - 4ac = 1600 > 0, it has real roots, so from equation 1 and equation 2, we
have,
x = [- b ± √(b2 - 4ac)]/2a
x = [- (0) ± √1600]/2
x = (0 ± 40)/2 
7) So,  x = (40)/2 or x = (- 40)/2, i.e. x = 20, or x = - 20.
8) As length is always positive, ignore x = - 20.
9) So, the breadth of the rectangular mango grove is 20 m and its length is 40 m.

Q4. Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their age in years was 48.

1) Let the present age of the first friend be x.
2) So, the present age of the second friend will be (20 - x).
3) 4 years ago, their ages will be (x - 4) and (20 - x - 4).
4) According to the problem,
(x - 4)(16 - x) = 48
x(16 - x) - 4(16 - x) = 48
16x - x2 - 64 + 4x = 48
20x - x2 - 64 - 48 = 0
20x - x2 - 112 = 0
x2 - 20x + 112 = 0 ------------------ equation 1.
5) Equate the coefficient of equation x2 - 20x + 112 = 0 with ax2 + bx + c = 0, so
a = 1, b = - 20, c = 112
6) First we will find:
b2 - 4ac = (- 20)2 - 4(1)(112)
b2 - 4ac = 400 - 448
b2 - 4ac = - 48 ------------------ equation 2.
7) As b2 - 4ac = - 48 < 0, it has no real roots, so the given situation is impossible.

Q5. Is it possible to design a rectangular park of perimeter 80 m and an area
of 400 m2? If so, find its length and breadth.

1) Let the breadth of a rectangular park be x m.
2) As, the perimeter of a rectangular park = 80 m.
3) So, length = [(perimeter/2) - breadth]
length = [(80/2) - x]
length = (40 - x)
4) According to the problem,
x(40 - x) = 400
40x - x2 = 400
x2 - 40x + 400 = 0 ------------------ equation 1.
5) Equate the coefficient of equation x2 - 40x + 400 = 0 with ax2 + bx + c = 0, so
a = 1, b = - 40, c = 400
6) First we will find:
b2 - 4ac = (- 40)2 - 4(1)(400)
b2 - 4ac = 1600 - 1600
b2 - 4ac = 0 ------------------ equation 2.
7) As, b2 - 4ac = 0, it has two equal real roots, so from equation 1 and equation 2,
we have,
x = [- b ± √(b2 - 4ac)]/2a
x = [- (- 40) ± √0]/2(1)
x = (40 ± 0)/2
x = (20 ± 0)
8) So,  x = 20, the breadth = 20 m and the length = 40 - 20 = 20 m.

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