NCERT Solution for Math Class 10
1. 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 ₹ 15 for the first km and ₹8 for each additional km.
(ii) The amount of air present in a cylinder when a vaccum pump removes 1/4 of the air remaining in the cylinder at a time.
(iii) The cost of a digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.
Solutions :
(i) Taxi fare for 1 km = ₹15
Taxi fare for 2 km = ₹15 + ₹8 = ₹23
Taxi fare for 3 km = ₹23 + ₹8 = ₹31
∴ seris is, 15, 23, 31..................
Now,
Difference between IInd and Ist term = 23 - 15 = 8
Difference between IIIrd and IInd term = 31 - 23 = 8
Since, difference is same. So, it is an Arithmetic Progression.
(ii) Let, the volume of cylinder = 16 litres
Now, according to question,
Air removed by pump = 1/4 of 16 = 1/4 × 16 = 4 litres
∴ Air present after first removal = 16 - 4 = 12 litres
Now,
Air again removed by pump = 1/4 of 12 = 1/4 × 12 = 3 litres
∴ Air present after second removal = 12 - 3 = 9 litres
The amount of air present in the cylinder is the series 16, 12, 9.........
Difference between IInd and Ist term = 12 - 16 = -4
Difference between IIIrd and IInd Term = 9 - 12 = -3
Since, difference is not same. So, it is not an Arithmetic Progression.
(iii) Cost of digging first metre = ₹150
Cost of digging second metre = ₹150 + ₹50 = ₹200
Cost of digging third metre = ₹200 + ₹50 = ₹250
∴ series is 150, 200, 250..........
Difference between IInd and Ist term = 200 - 150 = 50
Difference between IIIrd and IInd term = 250 - 200 = 50
Since, difference is same. So, it is an Arithmetic Progression.
(iv) Original amount = ₹10,000
For compound interest, we use the formula
Difference between IInd and Ist Term = 11664 - 10800 = 864
Difference between IIIrd and IInd Term = 12597.12 - 11664 = 933.12
Since, difference is not same. So, it is not an Arithmetic Progression.
2. Write 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
Solution :
First term, a = 10
Common difference, d = 10
Second term = First Term + Common Difference
= 10 + 10
= 20
Third term = second term + common difference
= 20 + 10
= 30
Fourth Term = Third Term + Common Difference
= 30 + 10
= 40
Hence, first four terms of AP are 10, 20, 30 and 40
(ii) a = -2, d = 0
Solution :
First Term, a = -2
Common Difference, d = 0
Second Term = First Term + Common Difference
= -2 + 0
= -2
Third Term = second term + common difference
= -2 + 0
= -2
Fourth Term = Third Term + Common Difference
= -2 + 0
= -2
Hence, first four term of AP are -2, -2, -2 and -2
(iii) a = 4, d = -3
Solution :
First Term, a = 4
Common Difference, d = -3
Second Term = First Term + Common Difference
= 4 + (-3)
= 4 - 3
= 1
Third Term = Second Term + Common Difference
= 1 + (-3)
= 1 - 3
= -2
Fourth Term = Third Term + Common Difference
= -2 + (-3)
= -2 - 3
= -5
Hence, first four term of AP are 4, 1, -2 and -5
(iv) a = -1, d = 1/2
Solution :
First term, a = -1
Common difference, d = 1/2
Second Term = First Term + Common Difference
= -1 + 1/2
= -1/2
Third Term = Second Term + Common Difference
= -1/2 + 1/2
= 0
Fourth Term = Third Term + Common Difference
= 0 + 1/2
= 1/2
Hence, first four term of AP are -1, -1/2, 0 and 1/2
(v) a = -1.25, d = -0.25
Solution :
First term, a = -1.25
Common difference, d = -0.25
Second Term = First Term + Common Difference
= -1.25 + (-0.25)
= -1.25 - 0.25
= -1.5
Third Term = Second Term + Common Difference
= -1.5 + (-0.25)
= -1.5 - 0.25
= -1.75
Fourth Term = Third Term + Common Difference
= -1.75 + (-0.25)
= -1.75 - 0.25
= -2
Hence, first four terms of AP are -1.25, -1.5, -1.75 and -2
3. For the following APs, write the first term and common difference :
(i) 3, 1, -1, -3,....
Solution :
Let, a be the first term and d be the common difference. Then,
a = 3
d = Second Term - First Term
= 1 - 3
= -2
(ii) -5, -1, 3, 7,...
Solution :
Let a be the first term and d be the common difference, then
a = -5
d = Seond term - First term
= -1 - (-5)
= -1 + 5
= 4
(iii) 1/3, 5/3, 9/3, 13/3,...
Solution :
Let a be the first term and d be the common difference, then
a = 1/3
d = second term - first term
= 5/3 - 1/3
= 4/3
(iv) 0.6, 1.7, 2.8, 3.9,...
Solution :
Let a be the first term and d be the common difference, then
a = 0.6
d = second term - first term
= 1.7 - 0.6
= 1.1
4. 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,...
Solution :
Second term - first term = 4 - 2 = 2
Third term - second term = 8 - 4 = 4
second term - first ≠ third term - second term
Thus, given sequence is not an AP.
(ii) 2, 5/2, 3, 7/2,...
Solution :
Second term - first term = 5/2 - 2 = 1/2
third term - second term = 3 - 5/2 = 1/2
second term - first term = third term - second term
Thus, the given sequence is an AP.
first term = 2
common difference = 1/2
Next three terms are,
fifth term = fourth term + commom difference = 7/2 + 1/2 = 8/2 = 4
sixth term = fifth term + common difference = 4 + 1/2 = 9/2
seventh term = sixth term + common difference = 9/2 + 1/2 = 10/2 = 5
(iii) -1.2, -3.2, -5.2, -7.2,...
Solution :
second term - first term = -3.2 - (-1.2) = -3.2 + 1.2 = -2
third term - second term = -5.2 - (-3.2) = -5.2 + 3.2 = -2
second term - first term = third term - second term
Thus, the given sequence is an AP.
first term = -1.2
common difference = -2
Next three terms are,
fifth term = fourth term + common difference = -7.2 + (-2) = -7.2 - 2 = -9.2
sixth term = fifth term + common difference = -9.2 + (-2) = -9.2 - 2 = -11.2
seventh term = sixth term + common difference = -11.2 + (-2) = -11.2 - 2 = -13.2
(iv) -10, -6, -2, 2,...
Solution :
second term - first term = -6 - (-10) = -6 + 10 = 4
third term - second term = -2 - (-6) = -2 + 6 = 4
second term - first term = third term - second term
Thus, the given sequence is an AP.
first term = -10
common difference = 4
Next three terms are ,
fifth term = fourth term + common difference = 2 + 4 = 6
sixth term = fifth term + common difference = 6 + 4 = 10
seventh term = sixth term + common difference = 10 + 4 = 14
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2,...
Solution :
second term - first term = 3 + √2 - 3 = √2
third term - second term = 3 + 2√2 - (3 + √2) = 3 + 2√2 - 3 - √2 = √2
Thus, the given sequence is an AP.
first term = 3
common difference = √2
Next three terms are,
fifth term = fourth term + common difference = 3 + 3√2 + √2 = 3 + 4√2
sixth term = fifth term + common difference = 3 + 4√2 + √2 = 3 + 5√2
seventh term = sixth term + common difference = 3 + 5√2 + √2 = 3 + 6√2
(vi) 0.2, 0.22, 0.222, 0.2222,...
Solution :
second term - first term = 0.22 - 0.2 = 0.02
third term - second term = 0.222 - 0.22 = 0.002
second term - first term ≠ third term - second term
Thus, the given sequence is not an AP.
(vii) 0, -4, -8, -12,...
Solution :
second term - first term = -4 -0 = -4
third term - second term = -8 - (-4) = -8 + 4 = -4
Thus, the given sequence is an AP.
first term = 0
common difference = -4
fifth term = fourth term + common difference = -12 + (-4) = -12 - 4 = -16
sixth term = fifth term + common difference = -16 + (-4) = -16 - 4 = -20
seventh term = sixth term + common difference = -20 + (-4) = -20 - 4 = -24
(viii) -1/2, -1/2, -1/2, -1/2,...
Solution :
second term - first term = -1/2 - (-1/2) = -1/2 + 1/2 = 0
third term - second term = -1/2 - (-1/2) = -1/2 + 1/2 = 0
second term - first term = third term - second term
Thus, the given sequence is an AP.
first term = -1/2
common difference = 0
Next three terms are,
fifth term = fourth term + common difference = -1/2 + 0 = -1/2
sixth term = fifth term + common difference = -1/2 + 0 = -1/2
seventh term = sixth term + common difference = -1/2 + 0 = -1/2
(ix) 1, 3, 9, 27,...
Solution :
second term - first term = 3 - 1 = 2
third term - second term = 9 - 3 = 6
second term - first term ≠ third term - second term
Thus, the given sequence is not an AP.
(x) a, 2a, 3a, 4a,...
Solution :
second term - first term = 2a - a = a
third term - second term = 3a - 2a = a
second term - first term = third term - second term
Thus, the given sequence is an AP.
first term = a
common difference = a
Next three terms are :
fifth term = fourth term + common difference = 4a + a = 5a
sixth term = fifth term + common difference = 5a + a = 6a
seventh term = sixth term + common difference = 6a + a = 7a
(xii) √2, √8, √18, √32,...
Solution :
second term - first term = √8 - √2 = 2√2 - √2 = √2
third term - second term = √18 - √8 = 3√2 - 2√ 2 = √2
second term - first term = third term - second term
Thus, the given sequence is an AP.
First term = √2
Common difference = √2
Next three terms are :
fifth term = fourth term + common difference = √32 + √2 = 4√2 + √2 = 5√2
sixth term = fifth term + common difference = 5√2 + √2 = 6√2
(xiii) √3, √6, √9, √12,...
Solution :
second term - first term = √6 - √3
third term - second term = √9 - √6 = 3 - √6
second term - first term ≠ third term - second term
Hence, It is not an AP.
.png "NCERT Solution for Math Class 10 Ch 5 AP")
