Class10 Maths Exercise 5.1 – Q(2) Q(3) Q(4)
NCERT Class10 maths Chapter 5 Exercise 5.1
Class10 Maths Exercise 5.1 – In this Question, I have explained Arithmetic Progression Concepts. Some series are AP or not is explained in the video and Written solution of Chapter 5 Exercise 5.1 Q(2) . Class 10 maths chapter 5 AP solutions in pdf form, video form and word form is here.
Class 10 maths chapter 5 exercise 5.1 Question – 1
CLASS 10 MATHEMATICS
CH-5 ARITHMETIC PROGRESSION (AP)
EX-5.1
Q2. Write first four terms of the A.P. when the first term a and the common difference are given as follows:
(i) a = 10, d = 10
ANSWER:
a = 10, d = 10
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
a1 = a = 10
a2 = a1+d = 10+10 = 20
a3 = a2+d = 20+10 = 30
a4 = a3+d = 30+10 = 40
a5 = a4+d = 40+10 = 50
THUS,
Therefore, the A.P. series will be 10, 20, 30, 40, 50 …
And First four terms of this A.P. will be 10, 20, 30, and 40.
(ii) a = – 2, d = 0
ANSWER:
Let the AP be a1, a2, a3, a4, a5 …
a1 = a = -2
a2 = a1+d = – 2+0 = – 2
a3 = a2+d = – 2+0 = – 2
a4 = a3+d = – 2+0 = – 2
Therefore, the A.P. series will be – 2, – 2, – 2, – 2 …
THUS,
First four terms of this A.P. will be – 2, – 2, – 2 and – 2.
(iii) a = 4, d = – 3
ANSWER:
Let the AP be a1, a2, a3, a4, a5 …
a1 = a = 4
a2 = a1+d = 4-3 = 1
a3 = a2+d = 1-3 = – 2
a4 = a3+d = -2-3 = – 5
Therefore, the A.P. series will be 4, 1, – 2 – 5 …
THUS,
first four terms of this A.P. will be 4, 1, – 2 and – 5.
(iv) a = – 1, d = 1/2
ANSWER:
Let the AP be a1, a2, a3, a4, a5 …
a2 = a1+d = -1+1/2 = -1/2
a3 = a2+d = -1/2+1/2 = 0
a4 = a3+d = 0+1/2 = 1/2
Therefore, the A.P. series will be-1, -1/2, 0, 1/2
THUS, First four terms of this A.P. will be -1, -1/2, 0 and 1/2.
(v) a = – 1.25, d = – 0.25
ANSWER
:
Let the AP be a1, a2, a3, a4, a5 …
a1 = a = – 1.25
a2 = a1 + d = – 1.25-0.25 = – 1.50
a3 = a2 + d = – 1.50-0.25 = – 1.75
a4 = a3 + d = – 1.75-0.25 = – 2.00
Therefore, the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 ……..
THUS, first four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.
CLASS 10 MATHEMATICS
CH-5 ARITHMETIC PROGRESSION (AP)
EX-5.1
Q3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, – 1, – 3 …
ANSWER:
Given series;
3, 1, – 1, – 3 …
First term, a = 3
Common difference, d = a2-a1 or
an – an-1
therefore,
d=1 – 3 = -2
THUS, d = -2
(ii) Given series, – 5, – 1, 3, 7 …
ANSWER:
First term, a = -5
Common difference, d = a2-a1
therefore,
d= ( – 1)-( – 5) = – 1+5 = 4
THUS, d=4
(iii) Given series, 1/3, 5/3, 9/3, 13/3 ….
ANSWER:
First term, a = 1/3
Common difference, d = a2-a1
therefore,
d= 5/3 – 1/3 = 4/3
THUS,
d=4/3
(iv) Given series, 0.6, 1.7, 2.8, 3.9 …
ANSWER:
First term, a = 0.6
Common difference, d = a2-a1
therefore,
d=1.7 – 0.6=1.1
THUS,
d=1.1
CLASS 10 MATHEMATICS
CH-5 ARITHMETIC PROGRESSION (AP)
EX-5.1
Q4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) 2, 4, 8, 16 …
ANSWER:
given,
2, 4, 8, 16 …
Here, the common difference is;
a2 – a1 = 4 – 2 = 2
a3 – a2 = 8 – 4 = 4
a4 – a3 = 16 – 8 = 8
Since, an+1 – an or the common difference is not the same every time.
Therefore, the given series are not forming an A.P.
(ii) 2, 5/2, 3, 7/2 ….
ANSWER:
Given, 2, 5/2, 3, 7/2 ….
Here,
a2 – a1 = 5/2-2 = 1/2
a3 – a2 = 3-5/2 = 1/2
a4 – a3 = 7/2-3 = 1/2
Since, an+1 – an or the common difference is same every time.
Therefore, d = 1/2 and the given series are in A.P.
The next three terms are;
a5 = 7/2+1/2 = 4
a6 = 4 +1/2 = 9/2
a7 = 9/2 +1/2 = 5
(iii) -1.2, -3.2, -5.2, -7.2 …
ANSWER:
Given
, -1.2, – 3.2, -5.2, -7.2 …
Here,
a2 – a1 = (-3.2)-(-1.2) = -2
a3 – a2 = (-5.2)-(-3.2) = -2
a4 – a3 = (-7.2)-(-5.2) = -2
Since, an+1 – an or common difference is same every time.
Therefore, d = -2 and the given series are in A.P.
Hence, next three terms are;
a5 = – 7.2-2 = -9.2
a6 = – 9.2-2 = – 11.2
a7 = – 11.2-2 = – 13.2
(iv) -10, – 6, – 2, 2 …
ANSWER:
given,
-10,-6,-2,2
Here, the terms and their difference are;
a2 – a1 = (-6)-(-10) = 4
a3 – a2 = (-2)-(-6) = 4
a4 – a3 = (2 -(-2) = 4
Since, an+1 – an or the common difference is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Hence, next three terms are;
a5 = 2+4 = 6
a6 = 6+4 = 10
a7 = 10+4 = 14
(v) 3, 3+√2, 3+2√2, 3+3√2
ANSWER:
GIVEN,
3, 3+√2, 3+2√2, 3+3√2
Here,
a2 – a1 = 3+√2-3 = √2
a3 – a2 = (3+2√2)-(3+√2) = √2
a4 – a3 = (3+3√2) – (3+2√2) = √2
Since, an+1 – an or the common difference is same every time.
Therefore, d = √2 and the given series forms a A.P.
Hence, next three terms are;
a5 = (3+√2) +√2 = 3+4√2
a6 = (3+4√2)+√2 = 3+5√2
a7 = (3+5√2)+√2 = 3+6√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
ANSWER:
GIVEN,
3, 3+√2, 3+2√2, 3+3√2
Here,
a2 – a1 = 0.22-0.2 = 0.02
a3 – a2 = 0.222-0.22 = 0.002
a4 – a3 = 0.2222-0.222 = 0.0002
Since, an+1 – an or the common difference is not same every time.
Therefore, and the given series doesn’t forms a A.P.
(vii) 0, -4, -8, -12 …
ANSWER:
GIVEN,0, -4, -8, -12 …
Here,
a2 – a1 = (-4)-0 = -4
a3 – a2 = (-8)-(-4) = -4
a4 – a3 = (-12)-(-8) = -4
Since, an+1 – an or the common difference is same every time.
Therefore, d = -4 and the given series forms a A.P.
Hence, next three terms are;
a5 = -12-4 = -16
a6 = -16-4 = -20
a7 = -20-4 = -24
(viii) -1/2, -1/2, -1/2, -1/2 ….
ANSWER:
GIVEN, -1/2, -1/2, -1/2, -1/2 ….
Here,
a2 – a1 = (-1/2) – (-1/2) = 0
a3 – a2 = (-1/2) – (-1/2) = 0
a4 – a3 = (-1/2) – (-1/2) = 0
Since, an+1 – an or the common difference is same every time.
Therefore, d = 0 and the given series forms a A.P.
Hence, next three terms are;
a5 = (-1/2)-0 = -1/2
a6 = (-1/2)-0 = -1/2
a7 = (-1/2)-0 = -1/2
(ix) 1, 3, 9, 27 …
ANSWER:
GIVEN,1, 3, 9, 27 …
Here,
a2 – a1 = 3-1 = 2
a3 – a2 = 9-3 = 6
a4 – a3 = 27-9 = 18
Since, an+1 – an or the common difference is not same every time.
Therefore, and the given series doesn’t form a A.P.
(x) a, 2a, 3a, 4a …
ANSWER:
GIVEN,a, 2a, 3a, 4a …
Here,
a2 – a1 = 2a–a = a
a3 – a2 = 3a-2a = a
a4 – a3 = 4a-3a = a
Since, an+1 – an or the common difference is same every time.
Therefore, d = a and the given series forms a A.P.
Hence, next three terms are;
a5 = 4a+a = 5a
a6 = 5a+a = 6a
a7 = 6a+a = 7a
(xi) a, a2, a3, a4 …
ANSWER:
GIVEN,a, a2, a3, a4 …
Here,
a2 – a1 = a2–a = a(a-1)
a3 – a2 = a3 – a2 = a2(a-1)
a4 – a3 = a4 – a3 = a3(a-1)
Since, an+1 – an or the common difference is not same every time.
Therefore, the given series doesn’t forms a A.P.
(xii) √2, √8, √18, √32 …
ANSWER:
GIVEN,√2, √8, √18, √32 …
Here,
a2 – a1 = √8-√2 = 2√2-√2 = √2
a3 – a2 = √18-√8 = 3√2-2√2 = √2
a4 – a3 = 4√2-3√2 = √2
Since, an+1 – an or the common difference is same every time.
Therefore, d = √2 and the given series forms a A.P.
Hence, next three terms are;
a5 = √32+√2 = 4√2+√2 = 5√2 = √50
a6 = 5√2+√2 = 6√2 = √72
a7 = 6√2+√2 = 7√2 = √98
(xiii) √3, √6, √9, √12 …
ANSWER:
GIVEN,√3, √6, √9, √12 …
Here,
a2 – a1 = √6-√3 = √3×√2-√3 = √3(√2-1)
a3 – a2 = √9-√6 = 3-√6 = √3(√3-√2)
a4 – a3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)
Since, an+1 – an or the common difference is not same every time.
Therefore, the given series doesn’t form a A.P.
(xiv) 12, 32, 52, 72 …
ANSWER:
GIVEN,12, 32, 52, 72 …
Here,
a2 − a1 = 32−12 = 20
a3 − a2 = 52−32 = 20
a4 − a3 = 72−52= 20
Since, an – an-1 or the common difference is same every time.
Therefore, the given series forms a A.P.
The next three terms are
a5=a4+d=92
a6=a5+d=112
a7=a6+d=132
12,32,52,72,92, 112 ,132 are in AP
(xv) 12, 52, 72, 73 …
ANSWER:
Given, 12, 52,72,73….
Here,
a2 − a1 = 52−12 = 40
a3 − a2 = 72−52= 20
a4 − a3 = 73−72 = 1
Since, an–an-1 or the common difference isn’t same every time.
Therefore, the series isn’t AP
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