Q. 44.2( 196 Votes )

Which of the following are APs ? If they form an AP, find the common difference d andwrite 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 + root 2 , 3+2 root 2 , 3+3 root 2 , l ..

(vi) 0.2 , 0.22 , 0.222 , 0.2222 ,

(vii)0 ,-4 ,-8 ,-12 ,

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

(ix) 1, 3, 9, 27,…

(x) a_12a_13a_14a_1

(xi) a_1a^2 , a^3 , a^4 ,

(xii)root 2 , root 8 , root 18 , root 32 ,

(xiii)root 3 , root 6 , root 9 , root 12 ,

(xiv) 1^2 , 3^2 , 5^2 , 7^2 ,

(xv) 12, 52, 72, 73.....

Answer :

For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of AP
  
(i) if ak+1 – ak is same for different values of k then the series is an AP.


We have, a1 = 2, a2 = 4, a3 = 8 and a4 = 16


a4 – a3 = 16 – 8 = 8


a3 – a= 8 – 4 = 4


a2 – a1= 4 – 2 = 2


Here, ak+1 – ak is not same for all values of k.


Hence, the given series is not an AP.


(ii) As per the question:


a1 = 2,


a2 = 5/2,


a3 = 3


And


a4 = 7/2


a4 – a3 = – 3= 1/2


a3 – a2 = 3 – = 1/2


a2 – a1 = – 2 = 1/2


Now, we can observe that ak+1 – ak is same for all values of k.


Hence, it is an AP.


And, the common difference = 1/2


Next three terms of this series are:


a5 = a + 4d


= 2 + 4* 1/2


= 4


a6 = a + 5d


= 2 + 5 * 1/2


=


a7 = a + 6d


= 2 + 6 * 1/2


= 5


Hence,


The next three terms of the AP are: 4, 9/2 and 5


(iii) a4 – a3 = - 7.2 + 5.2 = - 2


a3 – a2 = - 5.2 + 3.2 = - 2


a2 – a1 = - 3.2 + 1.2 = - 2


In this, ak+1 – ak is same for all values of k.


Hence, the given series is an AP.


Common difference = - 2


Next three terms of the series are:


a5 = a + 4d


= -1.2 + 4 × (-2)


= -1.2 - 8 = -9.2


a6 = a + 5d


= -1.2 + 5 × (-2)


= -1.2 - 10 = -11.2


a7 = a + 6d


= -1.2 + 6 × (-2)


= -1.2 - 12 = -13.2


Next three terms of AP are: - 9.2, - 11.2 and – 13.2


(iv)a4 – a3 = 2 + 2 = 4


a3 – a2 = - 2 + 6 = 4


a2 – a1 = - 6 + 10 = 4


Here, ak+1 – ak is same for all values of k


Hence, the given series is an AP.


Common difference = 4


Next three terms of the AP are:


a5 = a + 4d


= -10 + 4 × 4


= -10 + 16 = 6


a6 = a + 5d


= -10 + 5 × 4


= -10 + 20 = 10


a7 = a + 6d


= -10 + 6 × 4


= -10 + 24 = 14


Next three terms of AP are: 6, 10 and 14.


(v) a4 – a3 = 3 + 3√2 – 3 - 2√2 = √2


a3 – a2 = 3 + 2√2 – 3 - √2 = √2


a2 – a1 = 3 + √2 – 3 = √2


Here, ak+1 – ak is same for all values of k


Hence, the given series is an AP.


Common difference = √2


Next three terms of the AP are


a6 = a + 5d = 3 + 5√2


a7 = a + 6d = 3 + 6√2


Next three terms of AP are: 3 + 4√2, 3 + 5√2 and 3 + 6√2


(v)

a4 – a3 = 0.2222 – 0.222 = 0.0002


a3 – a2 = 0.222 – 0.22 = 0.002


a2 – a1 = 0.22 – 0.2 = 0.02


Here, ak+1 – ak is not same for all values of k


Hence, the given series is not an AP.

(vii) Here; a4 – a3 = - 12 + 8 = - 4


a3 – a2 = - 8 + 4 = - 4


a2 – a1 = - 4 – 0 = - 4


Since ak+1 – ak is same for all values of k.


Hence, this is an AP.


The next three terms can be calculated as follows:


a5 = a + 4d = 0 + 4(- 4) = - 16


a6 = a + 5d = 0 + 5(- 4) = - 20


a7 = a + 6d = 0 + 6(- 4) = - 24


Thus, next three terms are; - 16, - 20 and – 24


(viii) Here, it is clear that d = 0


Since ak+1 – ak is same for all values of k.


Hence, it is an AP.


The next three terms will be same, i.e. – 1/2


(ix) a4 – a3 = 27 – 9 = 18


a3 – a2 = 9 – 3 = 6


a2 – a1 = 3 – 1 = 2


Since ak+1 – ak is not same for all values of k.


Hence, it is not an AP.


(x) a4 – a3 = 4a – 3a = a


a3 – a2 = 3a – 2a = a


a2 – a1 = 2a – a = a


Since ak+1 – ak is same for all values of k.


Hence, it is an AP.


Next three terms are:


a5 = a + 4d = a + 4a = 5a


a6 = a + 5d = a + 5a = 6a


a7 = a + 6d = a + 6a = 7a


Next three terms are; 5a, 6a and 7a.


(xi) Here, the exponent is increasing in each subsequent term.

a4 = a4, a3 = a3, a2 = a2, a1 = a

a- a3 = a4 - a3

a3 - a2 = a3 - a2

since, the difference is not same,

Since ak+1 – ak is not same for all values of k.


Hence, it is not an AP.


(xii) Different terms of this AP can also be written as follows:


√2, 2√2, 3√2, 4√2, ………..


a4 – a3 = 4√2 - 3√2 = √2


a3 – a2 = 3√2 - 2√2 = √2


a2 – a1 = 2√2 - √2 = √2


Since ak+1 – ak is same for all values of k.


Hence, it is an AP.


Next three terms can be calculated as follows:


a5 = a + 4d = √2 + 4√2 = 5√2


a6 = a + 5d = √2 + 5√2 = 6√2


a7 = a + 6d = √2 + 6√2 = 7√2


Next three terms are; 5√2, 6√2 and 7√2


(xiii) a4 – a3 = √12 - √9 = 2√3 – 3                ( √12 =√ 2 x 2 x 3 = 2√3)


       a3 – a2 = √9 - √6 = 3 - √6


       a2 – a1 = √6 - √3


Since ak+1 – ak is not same for all values of k.


Hence, it is not an AP


(xiv) The given terms can be written as follows:


1, 9, 25, 49, …


Here, a4 – a3 = 49 – 25 = 24


a3 – a2 = 25 – 9 = 16


a2 – a1 = 9 – 1 = 8


Since ak+1 – ak is not same for all values of k.


Hence, it is not an AP.

(xv) 12, 52, 72, 73.....

a = 1

d = 52 - 1 = 25 - 1 = 24

d = 72 - 52 = 49 - 25 = 24

d = 74 - 49 = 24

As, common difference is same. The sequence is in A.P.

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