which of the following list of numbers form an A.P.? if so , write the next four terms. ( vi ) 3+√5, 3+2√5, 3+3√5
Answers
Answer:
3+4√5,3+5√5,3+6√5,3+7√5
i) 4,10,16,22,…
From the question, it is given that,
First-term a=4
Then, difference d
=10−4=6
16−10=6
22−16=6
Therefore, common difference d=6
Hence, the numbers are form A.P.
(ii) −2,2,−2,2,…
From the question, it is given that,
First term a=−2
Then, difference d=−2−2=−4
−2−2=−4
2−(−2)=2+2=4
Therefore, common difference d is not the same in the given numbers.
Hence, the numbers are not form A.P.
(iii) 2,4,8,16,…
From the question, it is given that,
First term a=2
Then, difference d=4−2=2
8−4=4
16−8=8
Therefore, common difference dis not the same in the given numbers.
Hence, the numbers are not form A.P.
(iv) 2,5/2,3,7/2,…
From the question, it is given that,
First term a=2
Then, difference d=5/2−2=(5−4)/2=1/2
3−5/2=(6−5)/2=1/2
7/2−3=(7−6)/2=1/2
Therefore, common difference d=1/2
Hence, the numbers are form A.P.
(v)−10,−6,−2,2,…
From the question, it is given that,
First term a=−10
Then, difference d
=−6−(−10)=−6+10=4
−2−(−6)=−2+6=4
2−(−2)=2+2=4
Therefore, the common difference d=4
Hence, the numbers are form A.P.
(vi) 1
2
,3
2
,5
2
,7
2
,…
From the question, it is given that,
First term a=1
2
=1
Then, difference d
=3
2
−1
2
=9−1=8
5
2
−3
2
=25−9=16
7
2
−5
2
=49−25=24
Therefore, the common difference d is not same in the given numbers.
Hence, the numbers are not form A.P.
(vii)1,3,9,27,…
From the question, it is given that,
First term a=1=1
Then, difference d=3−1=2
9−3=6
27−9=18
Therefore, common difference dis not same in the given numbers.
Hence, the numbers are not form A.P.
(viii)
2
,
8
,
18
,
32
,…
Given numbers can be written as,
2
,2
2
,3
2
,4
2
,…
From the question, it is given that,
First-term a=V2
Then, difference d=2
2
−
2
=
2
3
2
−2
2
=
2
4
2
−3
2
=
2
Therefore, common difference d=
2
Hence, the numbers are form A.P.
(ix ) a,2a,3a,4a,…
From the question, it is given that,
First term a=a
Then, difference d=2a−a=a
3a−2a=a
4a−3a=a
Therefore, common difference d=a
Hence, the numbers are form A.P.
(x) a,2a+1,3a+2,4a+3,…
From the question it is given that,
First term a=a
Then, difference d=(2a+1)−a=2a+1−a=a+1
(3a+2)−(2a+1)=3a+2−2a−1=a+1
(4a+3)−(3a+2)=4a+3−3a−2=a+1
Therefore, common difference d=a+1
Hence, the numbers are form A.P.