0
41

NCERT/CBSE MATHS Class 10 Ex-5.1 Q No.4 Solutions : Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.

Hy Friends Welcome On NCERT MATHS SOLUTIONS !! Today we are going to solve the Question : Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.of Exercise 5.1 Question No 4 Solutions (Arithmetic Progressions) of Class 10th, which will prove to be very helpful for you.

CBSE/NCERT MATHS Class 10 Ex-5.1 Q No. 4 Solutions

If you are a student of CBSE, today we are going to give you the CBSE / NCERT Chapter : Arithmetic Progressions Exercise 5.1 Question No – 4 Solutions . Hope you like this post about Class 10th Maths Solution.

Here we would like to tell you that there are other websites on which the students demanded the Solution of NCERT MATHS Chapters, we had promised them that we will soon send you the questions of Class 9, And we completed our promise to them.We asked the good knowledgeable teachers of mathematics and asked them to solve the issues of CBSE’s mathematical questions and also said that the answers to our questions are such that the students have no difficulty in solving all those questions.

Arithmetic Progressions CHAPTER : 5

n this progression, for a given series, the terms used are the first term, the common difference between the two terms and nth term.

Suppose, a1, a2, a3, ……………., an is an AP, then; the common difference “ d ” can be obtained as; In this progression, for a given series, the terms used are the first term, the common difference between the two terms and nth term.

a, a + d, a + 2d, a + 3d, a + 4d, ………. ,a + (n – 1) d

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 …
(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) aa2a3a4 …
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 12, 32, 52, 72 …
(xv) 12, 52, 72, 73 …

Class 10 Exercise 5.1 Question No. 4 Solutions

(i) 2, 4, 8, 16 …
Here,
a2 – a1 = 4 – 2 = 2
a3 – a2 = 8 – 4 = 4
a4 – a3 = 16 – 8 = 8
⇒ an+1 – an is not the same every time.

Therefore, the given numbers are forming an A.P.

(ii) 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
⇒ an+1 – an is same every time.
Therefore, d = 1/2 and the given numbers are in A.P.
Three more 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 …
Here,
a2 – a1 = ( -3.2) – ( -1.2) = -2
a3 – a2 = ( -5.2) – ( -3.2) = -2
a4 – a3 = ( -7.2) – ( -5.2) = -2
⇒ an+1 – an is same every time.
Therefore, d = -2 and the given numbers are in A.P.
Three more 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 …
Here,
a2 – a1 = (-6) – (-10) = 4
a3 – a2 = (-2) – (-6) = 4
a4 – a3 = (2) – (-2) = 4
⇒ an+1 – an is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Three more terms are
a5 = 2 + 4 = 6
a6 = 6 + 4 = 10
a7 = 10 + 4 = 14

(v) 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
⇒ an+1 – an is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more 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 ….
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
⇒ an+1 – an is not the same every time.

Therefore, the given numbers are forming an A.P.

(vii) 0, -4, -8, -12 …
Here,
a2 – a1 = (-4) – 0 = -4
a3 – a2 = (-8) – (-4) = -4
a4 – a3 = (-12) – (-8) = -4
⇒ an+1 – an is same every time.
Therefore, d = -4 and the given numbers are in A.P.
Three more terms are
a5 = -12 – 4 = -16
a6 = -16 – 4 = -20
a7 = -20 – 4 = -24

(viii) -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
⇒ an+1 – an is same every time.
Therefore, d = 0 and the given numbers are in A.P.
Three more 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 …
Here,
a2 – a1 = 3 – 1 = 2
a3 – a2 = 9 – 3 = 6
a4 – a3 = 27 – 9 = 18
⇒ an+1 – an is not the same every time.

Therefore, the given numbers are forming an A.P.

(x) a, 2a, 3a, 4a …
Here,
a2 – a1 = 2a – a
a3 – a2 = 3a – 2a = a
a4 – a3 = 4a – 3a = a
⇒ an+1 – an is same every time.
Therefore, d = a and the given numbers are in A.P.
Three more terms are
a5 = 4a + a = 5a
a6 = 5a = 6a
a7 = 6a + a = 7a

(xi) aa2a3a4 …
Here,
a2 – a1 = a– a = (a – 1)
a3 – a2 = a aa(a – 1)
a4 – a3 = a4 – aa3(a – 1)
⇒ an+1 – an is not the same every time.

Therefore, the given numbers are forming an A.P.

(xii) √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
⇒ an+1 – an is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more 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 …
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)
⇒ an+1 – an is not the same every time.

Therefore, the given numbers are forming an A.P.

(xiv) 12, 32, 52, 72 …

Or, 1, 9, 25, 49 …..
Here,
a2 − a1 = 9 − 1 = 8
a3 − a= 25 − 9 = 16
a4 − a3 = 49 − 25 = 24
⇒ an+1 – an is not the same every time.

Therefore, the given numbers are forming an A.P.

(xv) 12, 52, 72, 73 …
Or 1, 25, 49, 73 …
Here,
a2 − a1 = 25 − 1 = 24
a3 − a= 49 − 25 = 24
a4 − a3 = 73 − 49 = 24
i.e., ak+1 − ak is same every time.
⇒ an+1 – an is same every time.
Therefore, d = 24 and the given numbers are in A.P.
Three more terms are
a5 = 73+ 24 = 97
a6 = 97 + 24 = 121
a= 121 + 24 = 145